Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
CHAPTER 4
THE BINOMIAL AND NORMAL
PROBABILITY MODELS
TABLE OF CONTENTS
Page
1 The Binomial Probability Distribution
Conditions for a Binomial Experiment, Bernoulli Trials
Mean and Standard Deviation of a Binomial Random Variable
Using Excel and Table 1 to Calculate Binomial Probabilities
2 The Normal Probability Model
The Standard Normal Distribution
Standardizing Normal Distributions
Mensa Example
Tele-Evangelist Example
3 Approximating a Binomial Model with the Normal Distribution
112
112
112
113
119
119
125
126
127
127
Advertising Campaign and Dear Abby Examples
ESP Example
128
129
4 Fundamental Ideas From The Fourth Session
130
Six Probability Model Exercises
Answer Sketches for Six Probability Model Exercises
5 Table One: The Binomial Distribution
6 Table Two: The Normal Distribution
131
133
135
143
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 111
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
The Binomial Probability Distribution
The simplest data-gathering process is counting the number of times a certain event occurs. We can
reduce almost an endless variety of situations to this simple counting process.
Consumer preference and opinion polls (i.e., sample surveys) are conducted frequently in business.
Some recent examples include polls to determine the number of voters in favor of legalizing casino
gambling in a particular state, and the number of Americans in favor of government-controlled health
care.
The Binomial distribution is a discrete probability distribution that is extremely useful for describing many
phenomena. The random variable of interest that follows the Binomial distribution is the number of
successes obtained in a sample of n observations. The Binomial distribution can be applied to
numerous applications, such as:
•
•
•
•
What is the probability that in a sample of 20 tires of the same type none will be defective if 10% of
all such tires produced at a particular plant are defective?
What is the probability that red will come up 15 or more times in 20 spins of the American roulette
wheel (38 spaces)?
What is the probability that a student can pass (that is, get at least 60% correct on) a 30-question
multiple-choice exam (each question containing four choices) if the student guesses each question?
What is the probability that a particular stock will show an increase in its closing price on a daily
basis over the next ten (consecutive) trading sessions if stock market price changes really are
random?
Conditions Required for a Binomial Experiment
1. There is a set of n trials, which can be classified as either “successes” or “failures.”
2. The probability of a success, π, is constant over the n trials.
3. The outcome for any one trial is independent of the outcome for any other trial.
These first three conditions specify Bernoulli Trials.
4. The Binomial random variable X counts the number of “successes” in n trials.
Mean and Standard Deviation of the Binomial Probability Distribution
A random variable following the Binomial distribution is completely specified by the two parameters n
and π. Note that for a random variable X following the Binomial (n, π) distribution:
µ X = E[ X ] = nπ , and σ X = n π (1 − π )
We can obtain binomial probabilities using a Table (like Table 1) or Excel.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 112
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Using Excel to find Binomial Probabilities
To obtain the probability Pr(Y = k) for a Binomial random variable Y with parameters n and π using
Excel, use the function =BINOMDIST(k, n, π, 0). Suppose Y is a Binomial random variable with n =
5 and π = .50. To find Pr(Y = 0), enter =BINOMDIST(0, 5, 0.5, 0) in any empty cell and press
Enter to obtain the probability .03125. In this way, a complete Binomial table for n = 5 and π = .50
can be obtained:
k
0
1
2
3
4
5
Pr(Y = k) .03125
.15625 .31250 .31250 .15625 .03125
Sometimes, we will want to obtain a cumulative (“less than or equal to”) probability. For instance,
suppose we want to find the probability that Y ≤ 4. To do this, we can either
• add together the individual probabilities: Pr(Y ≤ 4) = Pr(Y = 0) + Pr(Y = 1) + Pr(Y = 2) + Pr(Y =
3) + Pr(Y = 4)
• OR adjust the Excel command and use the function BINOMDIST(k, n, π, 1). For instance, in this
setting, we’d use =BINOMDIST(4, 5, 0.5, 1).
The “1” in the last part of the command asks for a cumulative (“less than or equal to”)
probability instead of a simple (“equal to”) probability.
A complete cumulative probability table for the Binomial distribution with n = 5 and π = .50 is easily
generated from the information above:
k
0
1
2
3
4
5
Pr(Y ≤ k) .03125
.18750 .50000 .81250 .96875 1.00000
Using Table 1 to find Binomial Probabilities (see pp. 135-142)
To use the Table of Binomial Probabilities [Table 1], we first select the appropriate subtable for our n
and then the correct probability π. The table gives the probability distribution of X, the number of
successes, for a wide variety of small Binomial experiments. Note that values of π below .50 can be
read at the top of the block and values of x are read on the left. For values of π above .5, values of π
are read at the bottom, and values of x on the right.
Suppose n = 5 and π = .30. Here’s the n = 5 section of Table 1, pulled from the table on page 135.
Note the agreement of the shaded column with the Excel results.
n=5
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
3
4
5
.7738
.2036
.0214
.0011
.0000
.0000
.5905
.3281
.0729
.0081
.0005
.0000
.4437
.3915
.1382
.0244
.0022
.0001
.3277
.4096
.2048
.0512
.0064
.0003
.2373
.3955
.2637
.0879
.0146
.0010
.1681
.3602
.3087
.1323
.0284
.0024
.1160
.3124
.3364
.1811
.0488
.0053
.0778
.2592
.3456
.2304
.0768
.0102
.0503
.2059
.3369
.2757
.1128
.0185
.0313
.1563
.3125
.3125
.1563
.0313
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
5
4
3
2
1
0
k↑
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 113
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Example 4.1. A Simple Binomial Distribution Example
Let X be a Binomial random variable with n = 4 trials and probability of success π = .70. Let x be the
number of successes observed.
(a) Find Pr(x = 3).
(b) Find Pr(x < 2).
(c) Find Pr(x ≥ 2).
(d) What would be the modal (most common) outcome (number of successes)?
We could, for instance, draw a tree of the n = 4 trials, as shown on the next page. From the tree, we
can directly find the probability distribution of X. For instance, PX(0) = Pr(X = 0) = .0081, and PX(1) =
Pr(X = 1) = 4(.0189) = .0756, etc.
In Excel, we can use statements of the form =BINOMDIST(k, 4, 0.7, 0) to find the simple (“equal
to”) probability distribution for X. For instance,
• =BINOMDIST(0, 4, 0.7, 0) produces 0.0081
• =BINOMDIST(1, 4, 0.7, 0) produces 0.0756
• =BINOMDIST(2, 4, 0.7, 0) produces 0.2646, etc.
As before, we can find cumulative probabilities using =BINOMDIST(k, 4, 0.7, 1).
To use Table 1, we look at the n = 4 section of the table on page 135, and π = .70, which is
reproduced (and shaded) here:
n=4
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
3
4
.8145
.1715
.0135
.0005
.0000
.6561
.2916
.0486
.0036
.0001
.5220
.3685
.0975
.0115
.0005
.4096
.4096
.1536
.0256
.0016
.3164
.4219
.2109
.0469
.0039
.2401
.4116
.2646
.0756
.0081
.1785
.3845
.3105
.1115
.0150
.1296
.3456
.3456
.1536
.0256
.0915
.2995
.3675
.2005
.0410
.0625
.2500
.3750
.2500
.0625
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
4
3
2
1
0
k↑
Note that we must read up from the bottom (since π here is .70, not .30), we have
Number of successes, x
0
1
2
3
4
PX(x)
.0081 .0756 .2646 .4116 .2401
We can use Table 1 to solve the problems posed previously:
(a) The probability that x equals 3 is just PX(3) = Pr(X = 3) = .4116.
(b) Pr(X < 2) = Pr(X = 0) + Pr(X = 1) = .0081 + .0756 = .0837.
(c) Pr(X ≥ 2) = Pr(X = 2) + Pr(X = 3) + Pr(X = 4) = .2646 + .4116 + .2401 = .9163.
(d) There are five possible outcomes: 0, 1, 2, 3 or 4 successes. The outcome with the highest
probability is 3 successes: Pr(X = 3) = .4116, which is higher than Pr(X = 0) = .0081, Pr(X = 1) =
.0756, Pr(X = 2) = .2646, or Pr(X = 4) = .2401.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 114
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Trial One
Trial Two
Trial Three
Trial Four
S on 4th (.70)
X (# of S’s)
4
Probability
(.7) 4 = .2401
F on 4 th
(.30)
3
(.7) 3 (.3) 1 =
.1029
S on 4th (.70)
3
(.7) 3 (.3) 1 =
.1029
F on 4 th
(.30)
2
(.7) 2 (.3) 2 =
.0441
S on 4th (.70)
3
(.7) 3 (.3) 1 =
.1029
F on 4 th
(.30)
2
(.7) 2 (.3) 2 =
.0441
S on 4th (.70)
2
(.7) 2 (.3) 2 =
.0441
F on 4 th
(.30)
1
(.7) 1 (.3) 3 =
.0189
S on 4th (.70)
3
(.7) 3 (.3) 1 =
.1029
F on 4 th
(.30)
2
(.7) 2 (.3) 2 =
.0441
S on 4th (.70)
2
(.7) 2 (.3) 2 =
.0441
F on 4 th
(.30)
1
(.7) 1 (.3) 3 =
.0189
S on 4th (.70)
2
(.7) 2 (.3) 2 =
.0441
F on 4 th
(.30)
1
(.7) 1 (.3) 3 =
.0189
S on 4th (.70)
1
(.7) 1 (.3) 3 =
.0189
F on 4 th
(.30)
0
(.3) 4 = .0081
S on 3 rd (.70)
S on 2nd
(.70)
F on 3 rd
(.30)
Success (S) on
1 st Trial
(.70 =
probability)
S on 3rd (.70)
F on 2 nd
(.30)
F on 3 rd
(.30)
S on 3rd (.70)
S on 2nd
(.70)
F on 3 rd
(.30)
Failure (F) on
1 st Trial
S on 3rd (.70)
(.30 =
probability)
F on 2 nd
(.30)
F on 3 rd
(.30)
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 115
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Answering the Four Questions from the Start of the Chapter
•
What is the probability that in a sample of 20 tires of the same type none will be defective if 10% of
all such tires produced at a particular plant are defective?
Consider each tire inspection to be a Bernoulli trial, with the event “sampled tire is defective” defined as
a success. We need the probability of 0 successes in n = 20 trials, with π = Pr(success) = .10 The
relevant Table 1 section follows. The probability of k = 0 successes is .1216.
n = 20
k↓
0
•
π = .10
.1216
What is the probability that red will come up 15 or more times in 20 spins of the American roulette
wheel (38 spaces)?
Consider each spin of the wheel to be a Bernoulli trial, with the event “ball stops in red space” defined
as a success. Here, we want to know the probability of 15 or more successes in n = 20 trials. If each
of the 38 spaces is equally likely, then since there are 18 red spaces, the probability of a success is
9
π = 18
38 = 19 =.474 . Since Table 1 does not give exact probabilities for this value of π, let’s use Excel
to solve the problem.
As we have seen, the appropriate Excel statement to find the “equal to k” probability distribution for X
is of the form =BINOMDIST(k, n, π, 0). In this case, we have n = 20, π = 9/19, and k = 15, …, 20.
Probability Statement
Pr(X = 15)
Pr(X = 16)
Pr(X = 17)
Pr(X = 18)
Pr(X = 19)
Pr(X = 20)
Excel Statement (n = 20, π = 9/19)
=BINOMDIST(15, 20, 9/19, 0)
=BINOMDIST(16, 20, 9/19, 0)
=BINOMDIST(17, 20, 9/19, 0)
=BINOMDIST(18, 20, 9/19, 0)
=BINOMDIST(19, 20, 9/19, 0)
=BINOMDIST(20, 20, 9/19, 0)
Result
0.00849
0.00239
0.00051
0.00008
7.59E-06
3.23E-07
Summing these results, the probability of 15 or more “red” is approximately .0115
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 116
Statistics Pre-Term Program – Chapter Four
•
Some Special Probability Models
What is the probability that a student can pass (that is, get at least 60% correct on) a 30-question
multiple-choice exam (each question containing four choices) if the student guesses each question?
Consider each question to be a Bernoulli trial, with the event “answer is correct” defined as a success.
We then want to know the probability of at least 18 (30 × 60%) successes in n = 30 trials, with π =
Pr(success) = ¼ = .25. Note that Pr(X is at least 18) = Pr(X ≥ 18) = Pr(X = 18) + Pr(X = 19) + … +
Pr(X = 30) in this case. The relevant part of Table 1 is summarized below. The probability appears to
be 0 (to four decimal places).
n = 30
k↓
17
18
≥ 19
•
π = .25
.0002
.0000
.0000
What is the probability that a particular stock will show an increase in its closing price on a daily
basis over the next ten (consecutive) trading sessions if stock market price changes really are
random?
Consider each day’s price change (increase or decrease) to be a Bernoulli trial, with the event “stock
price increases” defined as a success. We assume that “price changes are random” means that the
probability of an increase equals the probability of a decrease and that there is no chance that the price
will stay the same. We then want to know the probability of 10 successes in n = 10 trials, with π =
Pr(increase) = ½ = 0.5.
From Table 1, we can read off the probability as 0.0010.
Example 4.2. Bad Seafood1
In a study, Consumer Reports (February 1992) found widespread contamination and mislabeling of
seafood in supermarkets in New York City and Chicago. The study revealing one alarming statistic:
40% of the swordfish pieces available for sale had a level of mercury above the Food and Drug
Administration (FDA) maximum amount. In a random sample of twelve swordfish pieces, find the
probability that:
a. Exactly five of the swordfish pieces have mercury levels above the FDA maximum.
b. At least ten swordfish pieces have a mercury level above the FDA maximum.
c. At most seven swordfish pieces have a mercury level above the FDA maximum.
1
Sincich, Exercise 5.22.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 117
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Example 4.3. Office Supply Bids
An office supply company currently holds 30% of the market for supplying suburban governments. This
share has been quite stable and there is no reason to think it will change. The company has three bids
outstanding, prepared by its standard procedure. Let Y be the number of the company’s bids that are
accepted.
a.
Find the probability distribution of Y.
b.
What assumptions did you make in answering part a.?
unreasonable?
c.
Find the expected value and variance of Y.
d.
It could be argued that if the company loses on the first bid, that signals that a competitor is
cutting prices and the company is more likely to lose the other bids as well. Similarly, if the
company wins on the first bid, that signals that competitors are trying to improve profit margins
and the company is more likely to win the other bids also. If this argument is correct, and the
two sides of the argument balance out to a 30% market share for the company, will the
expected value and variance of Y be increased or decreased compared to the values found in
part c.?
Are any of these assumptions
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 118
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
The Normal Probability Model
We now turn our attention to continuous probability density functions, which arise due to some
measuring process on a phenomenon of interest. When a mathematical expression is available to
represent some underlying continuous phenomenon, the probability that various values of the random
variable occur within certain ranges or intervals may be calculated. However, the exact probability of a
particular value from a continuous distribution is zero. A particular continuous distribution, the
Normal, or Gaussian, model is of vital importance.
1. Numerous continuous phenomena seem to follow it or can be approximated to it.
2. We can use it to approximate various discrete probability distributions and thereby avoid much
computational drudgery.
3. It provides the basis for classical statistical inference because of its relationship to the Central
Limit Theorem.
Crucial theoretical properties of the Normal distribution include:
1. It is bell-shaped and symmetrical in its appearance.
2. Its measures of central tendency (mean, median and mode) are all identical.
3. The Empirical Rule holds exactly for the Normal model.
The Normal probability density function is:
f Y ( y) =
1
− 1 ( y − µ )2 σ 2
e ( 2)
=
2πσ
1
2πσ
 ( y − µ )2 
−
2 
e  2σ 
The values µ and σ in the normal density function are in fact the mean and standard deviation of Y.
The Normal distribution is completely determined once the parameters µ and σ are specified. A whole
family of normal distributions exists, but one differs from another only in the location of its mean µ and in
the variance σ2 of its values.
The Standard Normal Distribution (Table 2)
Tables of normal curve areas (probabilities) are always given for the Standard Normal Distribution.
With a little effort, we can calculate any Normal probability based on just this one table.
•
•
The Standard Normal random variable Z has mean µZ = 0 and standard deviation σZ = 1. Like all
Normal distributions, the standard normal is symmetric around zero, so zero is the median (50th
percentile) of the distribution, so Pr(Z ≤ 0) = Pr(Z ≥ 0) = .5
Table 2 (pages 143-144) gives the probability that a Standard Normal distribution will be less than
z. Page 143 shows negative values of z, page 144 shows positive z values.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 119
Statistics Pre-Term Program – Chapter Four
•
Some Special Probability Models
Since the Z distribution is continuous, the probability that Z is exactly (to an infinite number of
decimal places) equal to any particular value x is essentially zero, so, for instance:
Pr(Z ≤ x) = Pr(Z < x) for all x, and similarly, Pr(Z ≥ y) = Pr(Z > y) for all y.
There are, in total, seven different forms of probability statements that we might need to solve for the
standard Normal model, all of which can be solved by working with the entries in Table 1 (pp. 143144). Here they are:
Scenario 1. Pr (Z < a) = Pr (Z ≤ a) where a is any negative number.
As an example, let’s find Pr (Z < -2.03). This setting is especially easy – we just read out the value
provided by the Table. In this case, we need the table value at –2.03, which can be found on page 143
(the relevant row of the table is shown below):
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
-2.00
.02275
.02222
.02169
.02118
.02068
.02018
.01970
.01923
.01876
.01831
-2.03
The table gives the probability of a standard Normal value being less than or equal to z, so Pr(Z ≤ 2.03) = Pr(Z < -2.03) = .02118. This number corresponds to the area of the shaded region on the
picture.
Scenario 2. Pr (Z < b) = Pr (Z ≤ b) where b is any positive number.
Now, let’s find Pr (Z < 1.48). As in Scenario 1., we can just read the value we need from the Table –
the row we need is on page 144, and is reproduced below.
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
1.40
.91924
.92073
.92220
.92364
.92507
.92647
.92785
.92922
.93056
.93189
So Pr(Z < 1.48) = Pr(Z ≤ 1.48) = .93056.
This number is the area of the shaded region in the next picture.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 120
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
1.48
Scenario 3. Pr (Z > c) = Pr (Z ≥ c) where c is any negative number.
We can simply use the complements law in this setting. If we want to find Pr(Z > -1.96), we find the
table value corresponding to –1.96, as shown below.
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
-1.90
.02872
.02807
.02743
.02680
.02619
.02559
.02500
.02442
.02385
.02330
Thus, Pr(Z < -1.96) = .025, and by the complements law, Pr(Z > -1.96) = 1 - .025 = .97500
Scenario 4. Pr (Z > d) = Pr (Z ≥ d) where d is any positive number.
Again, we can use the complements law. Suppose we want to find Pr(Z > 0.39):
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.30
.61791
.62172
.62552
.62930
.63307
.63683
.64058
.64431
.64803
.65173
From Table 1, Pr(Z < 0.39) = .65173. Thus, Pr(Z > 0.39) = 1 - .65173 = .34827
Scenario 5. Pr (e < Z < f) = Pr (e < Z ≤ f) = Pr (e ≤ Z < f) = Pr (e ≤ Z ≤ f) where e and f are both
negative numbers.
Suppose we want to find Pr(-1.82 < Z < -0.47). The relevant rows in the table are:
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
-1.80
.03593
.03515
.03438
.03362
.03288
.03216
.03144
.03074
.03005
.02938
Z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
-0.40
.34458
.34090
.33724
.33360
.32997
.32636
.32276
.31918
.31561
.31207
Pr(Z < -1.82) = .03438, and Pr(Z < -0.47) = .31918. Consider the picture below. The shaded
double arrow indicates the area we need to find:
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 121
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
.31918
.03438
-1.82
-0.47
Now, we see that in order to find the probability of falling in between –1.82 and –0.47, i.e. Pr(-1.82 ≤
Z ≤ -0.47), we need to subtract .03438 from .31918 – so the result is
Pr(-1.82 ≤ Z ≤ -0.47) = .31918 - .03438 = .28480
A General Rule for Using Table 2
If we want to find Pr(a ≤ Z ≤ b) where a and b are any numbers with a < b, we have the following
relationship:
Pr(a ≤ Z ≤ b) = Pr(Z ≤ b) – Pr(Z ≤ a), or, equivalently,
Pr(a < Z < b) = Pr(Z < b) – Pr(Z < a).
Scenario 6. Pr (e < Z < f) = Pr (e < Z ≤ f) = Pr (e ≤ Z < f) = Pr (e ≤ Z ≤ f) where e is negative and
f is a positive number.
Suppose we want to find Pr(-2.33 < Z < +1.96). First, we use Table 1 to get the values:
Z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
-2.30
.01072
.01044
.01017
.00990
.00964
.00939
.00914
.00889
.00866
.00842
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
1.90
.97128
.97193
.97257
.97320
.97381
.97441
.97500
.97558
.97615
.97670
From the General Rule above, we see that
Pr(-2.33 < Z < 1.96) = Pr(Z < 1.96) – Pr(Z < -2.33) = .97500 - .00990 = .96510
So there’s the answer. Just in case you’re not convinced that the rule works, here’s the graphical
evidence:
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 122
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
.97500
.00990
-2.33
1.96
Again, the shaded double arrow indicates the area we need – and we obtain the value by subtracting
.00990 from .97500 to get .96510.
Scenario 7. Pr (e < Z < f) = Pr (e < Z ≤ f) = Pr (e ≤ Z < f) = Pr (e ≤ Z ≤ f) where e and f are both
positive numbers.
Suppose we want to find Pr(0.93 < Z < 2.08). We first pull the appropriate Table 1 values.
Z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.90
.81594
.81859
.82121
.82381
.82639
.82894
.83147
.83398
.83646
.83891
Z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
2.00
.97725
.97778
.97831
.97882
.97932
.97982
.98030
.98077
.98124
.98169
From the General Rule above, we see that
Pr(0.93 < Z < 2.08) = Pr(Z < 2.08) – Pr(Z < 0.93) = .98124 - .82381 = .15743
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 123
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Example 4.4. Using the Standard Normal Distribution Table (Table 2)
Let Z be a standard Normal random variable. Find
•
k so that Pr(0 ≤ Z ≤ k) = .40
.500
.400
0
k
To find the Z score that cuts off the data as above, we look in the body of Table 2 for the closest value
to .90000, then use the Z score associated with that value. Here, we have
.89973
.90147
1.28
1.29
Thus, k = 1.28 (we usually don’t bother to interpolate…)
•
j so that Pr(-j ≤ Z ≤ j) = .60
.30000
-j
.30000
0
j
To find the Z score that cuts off the data as above, we look in the body of Table 2 for the closest value
to .80000 (why?), then use the Z score associated with that value. We have
.80234
.79955
0.84
0.85
Thus, j = 0.84, since it is closer to the desired probability.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 124
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Standardizing the Normal Distribution
By using the transformation formula Z =
X − µX
, any Normal random variable X (with mean µX and
σX
standard deviation σX) is converted to a standardized Normal random variable Z, which is Normal with
mean 0 and standard deviation 1. Thus, we only need one table to find probabilities for all Normal
random variables.
For a given value of X, the corresponding value of Z, sometimes called a z-score, is the number of
standard deviations σX that X lies away from µX.
Example 4.5. Using Table 2 for Unstandardized Normal probabilities
If X is a Normal random variable with µ = 100 and σ = 20, an X value of 130 lies 1.5 standard
deviations above (to the right of) the mean µ and the corresponding z-score is 1.50. An X value of 85
is .75 standard deviations below the mean since
z=
1.
85 − 100
= −0.75
20
Find Pr(85 ≤ X ≤ 130).
130 − 100 
 85 − 100
Pr(85 ≤ X ≤ 130) = Pr 
≤Z≤

 20
20 
= Pr ( − 0.75 ≤ Z ≤ 150
. )
So we now need to find the sum of areas A and B in the following picture of the Standard Normal (Z)
distribution:
Area
B
- 0.75
Area
A
0
1.50
From our general rule, Pr(in Area A or Area B) = Pr(Z < 1.50) – Pr(Z < -0.75),
From Table 2, we have Pr(Z < 1.50) = .93319 and Pr(Z < -0.75) = .22663
So the difference (and our answer) is .93319 - .22663 = .70656.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 125
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
2.
Find Pr(80 ≤ X).
 80 − 100

Pr(80 ≤ X ) = Pr 
≤ Z
 20

= Pr( − 2.00 ≤ Z )
So we need to find the area marked by the shaded arrow in the following picture:
.02275
- 2.00
We know that Pr(Z ≤ -2.00) = .02275 from Table 2, and so our area is 1 - .02275 = .97725
Example 4.6. Mensa and Standardized Test Scores
The Stanford-Binet IQ test is standardized that the IQ scores of the population will follow a Normal
distribution with mean 100, and standard deviation 16. Mensa is an organization that only allows
people to join if their IQs are in the top 2% of the population.
a. What is the lowest Stanford-Binet IQ that still makes one eligible to join Mensa?
Let X be IQ, then X has a Normal distribution with µ = 100 and σ = 16. We need to find k so that
k − 100 

Pr(X > k) = .02. Converting to Z scores, we have Pr Z >
 =.02 . Now, Pr(Z > 2.05) = .02,

16 
k − 100
since Pr(Z < 2.05) ≈ .98. Thus,
= 2.05 , so k = 132.8.
16
b. According to the model, what percentage of the population score below 84 on the StanfordBinet measure?
Let X be IQ, then X has a Normal distribution with µ = 100 and σ = 16. The question asks for the
probability of a score below 84, i.e. Pr(X < 84). Converting our probability statement to Z scores, we
84 − 100 

obtain Pr( X < 84) = Pr  Z <
 = Pr ( Z < −1.00) . From Table 2, Pr(Z < -1.00) = .15866.

16 
So a bit less than 16% of the population score below 84. Note the similarity to the empirical rule in this
setting. Since roughly 68% of the scores should be within one standard deviation of the mean, 32%
should be outside that range, half of which (or 16%) will be below it, and half (or 16%) above it.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 126
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
c. Mensa also allows members to qualify on the basis of certain standard tests. The GRE is
standardized to have a Normal population distribution with mean 497, and standard deviation
115. If you were to try to qualify on the basis of the GRE exam, what score would you need?
Let Y be GRE, then Y has a Normal distribution with µ = 497 and σ = 115. Again we must find k so
k − 497 
that Pr(Y > k) = .02. Converting, we have Pr Z >
 =.02 . From Table 2, Pr(Z > 2.05) = .02,

115 
k − 497
since Pr(Z < 2.05) ≈ .98. Thus,
= 2.05 , so k = 732.75.
115
Example 4.7. Tele-Evangelist Collections 2
Records for the past several years show that the amount of money collected daily by a prominent teleevangelist is Normally distributed with a mean (µ) of $2000 and a standard deviation (σ) of $500.
1. What is the chance that tomorrow’s donations will be less than $1500?
2. What is the probability that tomorrow’s donations will be between $2000 and $3000?
3. What is the chance that tomorrow’s donations will exceed $3000?
Approximating a Binomial with the Normal Distribution
The Normal distribution may be used to approximate probabilities for many other distributions. For
instance, we can approximate the Binomial distribution by a Normal distribution for certain values of n
and π. The approximation works well as long as both nπ and n(1 - π) are greater than 5.
The basic idea is to pretend that a Binomial random variable Y has a Normal distribution, using the
Binomial mean µ = n π, and standard deviation σ = nπ (1 − π ) .
For example, we can treat a Binomial random variable with n = 400 and π = .20 as approximately
normally distributed with
µ = 400(.20) = 80 and σ2 = 400(.20)(.80) = 64, so σ = 8.
To approximate Pr(Y > 96), we use Table 2 to obtain
[96 − 80] 

P(Y > 96 ) = P Z >
 = P( Z > 2.00 ) =.0228


8
2
Larsen, Marx and Cooil, Question 6.3.1.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 127
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Example 4.8. Advertising Campaign
An advertising campaign for a new product is targeted to make 20% of the adult population in a
metropolitan area aware of the product. After the campaign, a random sample of 400 adults in the
metropolitan area is obtained.
a. Find the approximate probability that 57 or fewer adults in the sample are aware of the
product.
X, the number of adults aware of the product, follows a Binomial distribution with n = 400, and π = .20
(assuming that the ad campaign worked). Thus, E(X) = µ = (400)(.2) = 80.0, and SD(X) = σ =
400(.2)(.8) = 8 . We use a Normal approximation for X, and we want to find Pr(X ≤ 57). Converting to Z
57 − 80 

Pr Z ≤
 = Pr(Z ≤ -2.88) = .00199

8 
b. Should the approximation in part a. be accurate?
scores, Pr(X ≤ 57) =
Sure. nπ = (400)(.20) = 80.0, and n(1-π) = (400)(.80) = 320.0, both much larger than 5.
c. The sample showed that 57 adults were aware of the product. The marketing manager
argues that the low awareness rate is a random fluke of the particular sample. Based on your
answer to part a., do you agree?
No. A similar calculation to part a. suggests that the probability of a result this bad (low) in terms of
awareness turns out to be about .002 if the ad campaign really has met its target.
Example 4.9. Dear Abby Example3
Dear Abby,
You wrote in your column that a woman is pregnant for 266 days. Who said so? I carried my baby for ten
months and five days, and there is no doubt about it because I know the exact date my baby was conceived. My
husband is in the Navy and it couldn’t have possibly been conceived any other time because I saw him only once for
an hour, and I didn’t see him again until the day before the baby was born.
I don’t drink or run around, and there is no way this baby isn’t his, so please print a retraction about the
266-day carrying time because otherwise I am in a lot of trouble.
San Diego Reader
Whether or not San Diego Reader is telling the truth is a judgment that lies beyond the scope of any
statistical analysis (other than maybe a paternity test!), but quantifying the plausibility of her story
doesn’t. According to the collective experience of generations of pediatricians, pregnancy durations, Y,
tend to be Normally distributed with µ = 266 days and σ = 16 days.
What do these figures imply about San Diego Reader’s credibility? Do you believe her? Consider both
a Normal probability model, as well as the Excel simulation results which follow…
3
Larsen, Marx and Cooil, Question 5.3.12.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 128
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Simulation of 100 pregnancy durations using Excel’s Data Analysis Tool for Random Number
Generation (Normal Distribution, with mean µ = 266 days, and σ = 16 days).
100 Simulated Pregnancy Durations from Excel (sorted)
234.37 238.40 239.88 241.02 241.50 241.90 243.72 244.89 246.35 246.47
248.95 249.61 251.61 251.96 251.98 252.28 252.55 252.92 252.98 253.06
253.78 254.35 254.41 254.63 255.94 255.95 257.14 257.58 258.54 259.32
260.01 260.07 260.70 261.12 261.14 261.39 261.86 262.03 262.11 262.49
262.65 262.66 262.92 263.20 263.51 263.67 263.91 263.94 263.95 264.04
264.22 265.86 266.27 266.37 266.46 268.45 268.86 268.86 269.58 269.68
270.79 271.03 271.03 271.32 271.65 271.73 272.01 272.42 272.63 273.06 273.22
273.61 275.10 275.35 275.81 275.99 277.59 277.83 279.57 279.77
281.26 281.34 281.86 282.10 282.24 282.63 283.17 284.63 285.19 286.08
288.31 288.54 289.86 289.92 291.44 293.01 294.09 294.52 294.75 306.03
Example 4.10. ESP and The Normal Approximation
Research in extrasensory perception has ranged from the slightly unconventional to the downright
bizarre. Beginning around 1910, though, experimenters moved out of seance parlors and into
laboratories, where they began setting up controlled studies that could be analyzed statistically. In
1938, Pratt and Woodruff designed an experiment that became a prototype for an entire generation of
ESP research.
The investigator and a subject sat at opposite ends of a table. Between them was a partition with a
large gap at the bottom. Five blank cards, visible to both participants, were placed side by side on the
table beneath the partition. On the subject’s side of the partition, one of the standard ESP symbols
(square, wavy lines, circle, star and plus) was hung over each of the blank cards. The experimenter
shuffled a deck of ESP cards, looked at the top one, and concentrated. The subject tried to guess the
card. If (s)he thought it was a circle, (s)he would point to the bland card on the table that was beneath
the circle card hanging on the subject’s side of the partition. The procedure was then repeated.
Altogether, 32 subjects, all students, took part in the study. They made 60,000 guesses and were
correct 12,489 times.
What can we deduce from these numbers?
Have Pratt and Woodruff presented convincing evidence in favor of ESP, or should 12,489
right answers out of 60,000 attempts be considered inconclusive?
X follows a Binomial distribution, with n = 60,000 and π = .20. Since we don’t have a Binomial table
for n = 60,000, we’ll use a Normal approximation. It should work well since nπ = (60,000)(.20) =
12,000, and n(1-π) = (60,000)(.80) = 48,000 are both much larger than 5. To use the Normal
approximation, we need the mean and standard deviation of X. In this case, we have µ = nπ =
(60,000)(.20) = 12,000, and σ = (60000)(.2)(.8) = 97.98 .
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 129
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Fundamental Ideas From The Fourth Session
The Binomial Probability Distribution
•
•
Bernoulli Trials
• There is a set of n dichotomous (0/1 or “success” / “failure”) trials.
• The probability of a success, π, is constant over the n trials.
• The outcome of a trial is independent of the outcome for any other trial.
Binomial r.v. X = number of “successes” in n Bernoulli trials, each with prob. π:
Mean µ X = E[ X ] = nπ , and SD σ X = n π (1 − π )
Using Excel to find Binomial Probabilities (See Bulkpack Chapter 4)
•
•
If X follows a Binomial distribution with parameters n and π, to find Pr(X = k), enter
=BINOMDIST(k, n, π, 0)
If X follows a Binomial distribution with parameters n and π, to find Pr(X ≤ k), enter
=BINOMDIST(k, n, π, 1)
Using Table 1 to find Binomial Probabilities
•
•
n = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 20, 25, 30, 50 and 100
and π = .05, .10, .15, …, .90, .95
The Normal Probability Distribution
•
•
•
•
•
•
•
•
Crucial properties
Bell-shaped and symmetrical distribution.
Mean, Median and Mode all identical.
Empirical Rule holds exactly.
Completely specified by Mean µ and Standard Deviation σ
Applies in many settings, thanks mainly to Central Limit Theorem (to come), and to use of Normal
approximation to other distributions (like the Binomial).
The Standard Normal Distribution
Standard Normal random variable Z has mean µZ = 0 and standard deviation σZ = 1.
Table 2 gives the Standard Normal distribution
• Table 2 gives probabilities between 0 and a positive number z.
Z scores convert Normal random variables to Standard Normal random variables
X − µX
Z=
σX
Using the Normal Distribution to Approximate Other Distributions
Find the mean µ and standard deviation σ of the desired distribution, then use Normal probabilities with
these parameters to approximate.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 130
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Six Probability Model Exercises (Binomial & Normal Models )
Answer Sketches follow the exercises
1.
[Keller, Warrack and Bartel, p. 182] A warehouse engages in acceptance sampling to determine if
it will accept or reject incoming lots of designer sunglasses, some of which invariably are
defective. Specifically, the warehouse has a policy of examining a sample of 50 sunglasses from
each lot and of accepting the lot only if the sample contains no more than four defective pairs.
What is the probability of a lot’s being accepted if, in fact, 5% of the sunglasses in the lot are
defective?
2.
[KWB, p. 178] A student majoring in accounting is trying to decide upon the number of firms to
which she should apply. Given her work experience, grades, and extracurricular activities, she
has been told by a placement counselor that she can expect to receive a job offer from 80% of
the firms to which she applies. Wanting to save time, the student applies to only five firms.
a)
b)
c)
3.
[KWB, p. 202] A firm’s marketing manager believes that total sales for the firm can be modeled
by using a Normal distribution, with a mean of $2.5 million and a standard deviation of
$300,000.
a)
b)
c)
d)
4.
Determine the probability that the student receives no offers.
What is the probability of five offers?
What is the chance of at most two offers?
What is the probability that the firm’s sales will exceed $3 Million?
What is the probability that the firm’s sales will fall within $150,000 of the expected
level of sales?
In order to cover fixed costs, the firm’s sales must exceed the break-even level of $1.8
million. What is the probability that sales will exceed the break-even level?
Find the sales level that has only a 9% chance of being exceeded next year.
[KWB, p. 202] Empirical studies have provided support for the belief that a common stock’s
annual rate of return is approximately Normally distributed. Suppose you have invested in the
stock of a company for which the annual return has an expected value of 16% and a standard
deviation of 10%.
a)
b)
c)
d)
Find the probability that your one-year return will exceed 30%.
Find the probability that your one-year return will be negative.
Suppose this company embarks on a new high-risk, but potentially highly profitable
venture. As a result, the return on the stock now has an expected value of 25% and a
standard deviation of 20%. Answer parts (a) and (b) in light of the revised estimates
regarding the stock’s return.
As an investor, would you approve of the company’s decision to embark on the new
venture?
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 131
Statistics Pre-Term Program – Chapter Four
5.
6.
Some Special Probability Models
[Carlson & Thorne, p. 218] A building owner purchases four new refrigerators, which are
guaranteed by the manufacturer. The probability that each refrigerator is defective is 0.20.
Define the random variable X = total number of defective refrigerators.
a)
Construct the probability distribution of X.
b)
What are the mean and variance of X ?
c)
The cost of repair, to honor the guarantee, consists of a fixed fee of $40 plus a cost of
$60 to repair each defective refrigerator. What is the average cost to repair these four
refrigerators? Note that the fixed fee of $40 is incurred only if there is at least one
defective refrigerator.
It is believed by certain parapsychologists that hypnosis can bring out a person’s ESP ability.
To test their theory, they hypnotize 15 volunteers and ask each to make 100 guesses of ESP
cards under conditions much like those of Pratt and Woodruff described in Example 4.F in the
Lecture Notes. A total of X = 326 correct responses were recorded. To what extent does the
326 support the hypnosis hypothesis? To answer the question, compute Pr(X ≥ 326).
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 132
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
Six Probability Model Exercises: Answer Sketches
1.
We have to find the probability of accepting a lot when in fact 5% of the sunglasses in the lot are
defective. We accept the lot if a sample of n = 50 sunglasses produces no more than 4
defective pairs. If we take a random sample from the lot, we should be able to assume
independence of sunglasses and a constant probability of defects. Thus, we have a Binomial
experiment. Our random variable is X = the number of defective sunglasses in our sample from
the lot. So we know that X follows a Binomial distribution with n = 50 and π = .05, and we
want to find the probability that X is no more than four. Again, we can use Table 1.
Pr(X is no more than 4) = Pr(X ≤ 4) = Pr(X=0) + Pr(X=1) + Pr(X=2) + Pr(X=3) + Pr(X=4)
= .0769 + .2025 + .2611 + .2199 + .1360 = .8964
Or, we could use Excel’s BINOMDIST function. In this case, we’d want the cumulative (“less
than or equal to”) for X = 4 in a Binomial with n = 50 trials, and probability of success π = .05,
so we use =BINOMDIST(4,50,0.05,1) and obtain the value 0.896383.
2.
The number of firms making offers, X, follows a Binomial distribution with n = 5, and π = .80.
So the probability that the student receives no offers is Pr(X = 0) = .0003, according to Table
1. The probability of five offers is Pr(X = 5) = .3277, and the probability of at most two offers
is Pr(X is at most 2) = Pr(X ≤ 2) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2) = .0003 + .0064 +
.0512 = .0579.
3.
Sales, Y, can be modeled with a Normal distribution with µ = $2,500,000 and σ = $300,000.
− 3, 000 , 000
a)
Pr(Y > $3,000,000) = Pr Z > 2, 500, 000
= Pr(Z > -1.67). From Table 2, we
300 , 000
[
]
have Pr(Z < -1.67) = .04746, so Pr(Z > -1.67) = 1 - .04746 = .95254
b)
To fall within 150,000 of the expected value, 2,500,000, we must find Pr (2,350,000 <
− 2 , 500, 000
2 , 650, 000− 2 ,500, 000
Y < 2,650,000) = Pr( 2, 350 , 000
<Z<
) = Pr(-0.50 < Z < 0.50).
300 ,000
300 , 000
From Table 2, we have Pr(Z < .50) = .69146 and Pr(Z < -.50) = .30854, so
Pr(2,350,000 < Y < 2,650,000) = Pr(Z < 0.50) – Pr(Z < -0.50) = .69146 - .30854 =
.38292
c)
[
Pr(Y > 1,800,000) = Pr Z >
1, 800, 000 − 2 ,500 ,000
300 , 000
] = Pr(Z > -2.33). From Table 2, we have
Pr(Z < -2.33) = .00990. So Pr(Y > 1,800,000) = Pr(Z > -2.33) = 1 - .00990 =
.99010
d)
We want to find the value x such that Pr(Y > x) = .0900. To do this, we first note that
2 ,500, 000
this means that we have Pr Z > x −300
= .09. Now, the value from Table 2 that cuts
, 000
[
]
off probability .09 in the upper tail of the standard Normal distribution is the value that
cuts off probability .91 in the lowe tail (why?). This value is approximately Z = 1.34,
since Pr(Z < 1.34) = .90988, so that Pr(Z > 1.34) = .09012. Thus, we need to solve
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 133
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
x − 2,500,000
the following equation for x:
= 134
. . So we have x - 2,500,000 =
300,000
402,000 and so x = $2,902,000, which is the sales level that has only a 9% chance of
being exceeded next year.
4.
5.
6.
Y, annual rate of return, follows a Normal distribution with mean 16 and standard deviation 10.
a)
Pr(Y > 30) = Pr(Z > [30-16]/10) = Pr(Z > 1.40). From Table 2, Pr(Z < 1.40) =
.91924, so Pr(Y > 30) = Pr(Z > 1.40) = 1 - .91924 = .08076
b)
Pr(Y > 0) = Pr(Z > [0-16]/10) = Pr(Z > -1.60). From Table 2, Pr(Z < -1.60) =
.05480, so Pr(Y > 0) = Pr(Z > -1.60) = 1 - .0548 = .94520
c)
[Revised part a] Pr(Y > 30) = Pr(Z > [30-25]/20) = Pr(Z > 0.25). From Table 2,
Pr(Z < 0.25) = .59871, so Pr(Y > 30) = Pr(Z > 0.25) = 1 - .59871 = .40129.
[Revised part b] Pr(Y > 0) = Pr(Z > [0-25]/20) = Pr(Z > -1.25). From Table 2, Pr(Z
< -1.25) = .10565, so Pr(Y > 0) = Pr(Z > -1.25) = 1 - .10565 = .89435.
d)
According to these calculations, the chance of earning 30% or more is increased, as is
the chance of a loss. Thus, your decision as to whether to support the decision depends
on how risk-averse you are. If you are very risk-adverse, you won’t support it. If you
are not, you might support it.
X follows a Binomial distribution with n = 4 and π = .20
a)
X, from Table 1, has the following probability distribution: Pr(X = 0) = .4096, Pr(X =
1) = .4096, Pr(X = 2) = .1536, Pr(X = 3) = .0256, Pr(X = 4) = .0016
b)
E(X) = nπ = (4)(.2) = 0.8, and Var(X) = n π (1 - π) = 4(.2)(.8) = 0.64
c)
We can either slog this out directly, or use the fact that Variable Cost = 60X, so
E(Variable Cost) = E(60X) = 60E(X) = 60(.8) = $48. Also, note that Fixed Cost =
40 if X ≥ 1. Pr(X ≥ 1) = 1 - Pr(X = 0) = .5904 so the expected fixed cost is
40(.5904) = $23.62. Thus, the expected total cost should be 23.62 + 48 = $71.62
X follows a Binomial distribution, with n = 1500 and π = .20. The Normal approximation
should work well since nπ = (1500)(.20) = 300, and n(1-π) = (1500)(.80) = 1200 are both
much larger than 5. Here, we have mean µ = nπ = (1500)(.20) = 300, and standard deviation
σ = (1500)(.2)(.8) = 15.49 . Converting our probability statement to Z scores, we have
326 − 300 

Pr(X ≥ 326) = Pr Z ≥
 = Pr(Z ≥ 1.68) = 1 - .95352 = .04648

15.49 
So the chances of observing a result this strongly or more in favor of the hypnosis hypothesis are
just a bit less than 1 in 20.
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 134
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
TABLE 1 – The Binomial Probability Distribution
Tabulated Values are Pr(X = k) [Values are rounded to four decimal places.]
Examples: For n = 2, π = .35, Pr(X = 0) = .4225; For n = 3, π = .60, Pr(X = 1) = .2880
n=2
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
.9025
.0950
.0025
.8100
.1800
.0100
.7225
.2550
.0225
.6400
.3200
.0400
.5625
.3750
.0625
.4900
.4200
.0900
.4225
.4550
.1225
.3600
.4800
.1600
.3025
.4950
.2025
.2500
.5000
.2500
2
1
0
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
k↑
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
3
.8574
.1354
.0071
.0001
.7290
.2430
.0270
.0010
.6141
.3251
.0574
.0034
.5120
.3840
.0960
.0080
.4219
.4219
.1406
.0156
.3430
.4410
.1890
.0270
.2746
.4436
.2389
.0429
.2160
.4320
.2880
.0640
.1664
.4084
.3341
.0911
.1250
.3750
.3750
.1250
3
2
1
0
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
k↑
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
3
4
.8145
.1715
.0135
.0005
.0000
.6561
.2916
.0486
.0036
.0001
.5220
.3685
.0975
.0115
.0005
.4096
.4096
.1536
.0256
.0016
.3164
.4219
.2109
.0469
.0039
.2401
.4116
.2646
.0756
.0081
.1785
.3845
.3105
.1115
.0150
.1296
.3456
.3456
.1536
.0256
.0915
.2995
.3675
.2005
.0410
.0625
.2500
.3750
.2500
.0625
4
3
2
1
0
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
k↑
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
3
4
5
.7738
.2036
.0214
.0011
.0000
.0000
.5905
.3281
.0729
.0081
.0005
.0000
.4437
.3915
.1382
.0244
.0022
.0001
.3277
.4096
.2048
.0512
.0064
.0003
.2373
.3955
.2637
.0879
.0146
.0010
.1681
.3602
.3087
.1323
.0284
.0024
.1160
.3124
.3364
.1811
.0488
.0053
.0778
.2592
.3456
.2304
.0768
.0102
.0503
.2059
.3369
.2757
.1128
.0185
.0313
.1563
.3125
.3125
.1563
.0313
5
4
3
2
1
0
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
k↑
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
3
4
5
6
.7351
.2321
.0305
.0021
.0001
.0000
.0000
.5314
.3543
.0984
.0146
.0012
.0001
.0000
.3771
.3993
.1762
.0415
.0055
.0004
.0000
.2621
.3932
.2458
.0819
.0154
.0015
.0001
.1780
.3560
.2966
.1318
.0330
.0044
.0002
.1176
.3025
.3241
.1852
.0595
.0102
.0007
.0754
.2437
.3280
.2355
.0951
.0205
.0018
.0467
.1866
.3110
.2765
.1382
.0369
.0041
.0277
.1359
.2780
.3032
.1861
.0609
.0083
.0156
.0938
.2344
.3125
.2344
.0938
.0156
6
5
4
3
2
1
0
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
k↑
n=3
n=4
n=5
n=6
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 135
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
TABLE 1 – The Binomial Probability Distribution
Tabulated Values are Pr(X = k) [Values are rounded to four decimal places.]
Examples: For n = 7, π = .15, Pr(X = 2) = .2097; For n = 8, π = .55, Pr(X = 3) = .1719
n=7
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
3
4
5
6
7
.6983
.2573
.0406
.0036
.0002
.0000
.0000
.0000
.4783
.3720
.1240
.0230
.0026
.0002
.0000
.0000
.3206
.3960
.2097
.0617
.0109
.0012
.0001
.0000
.2097
.3670
.2753
.1147
.0287
.0043
.0004
.0000
.1335
.3115
.3115
.1730
.0577
.0115
.0013
.0001
.0824
.2471
.3177
.2269
.0972
.0250
.0036
.0002
.0490
.1848
.2985
.2679
.1442
.0466
.0084
.0006
.0280
.1306
.2613
.2903
.1935
.0774
.0172
.0016
.0152
.0872
.2140
.2918
.2388
.1172
.0320
.0037
.0078
.0547
.1641
.2734
.2734
.1641
.0547
.0078
7
6
5
4
3
2
1
0
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
k↑
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
3
4
5
6
7
8
.6634
.2793
.0515
.0054
.0004
.0000
.0000
.0000
.0000
.4305
.3826
.1488
.0331
.0046
.0004
.0000
.0000
.0000
.2725
.3847
.2376
.0839
.0185
.0026
.0002
.0000
.0000
.1678
.3355
.2936
.1468
.0459
.0092
.0011
.0001
.0000
.1001
.2670
.3115
.2076
.0865
.0231
.0038
.0004
.0000
.0576
.1977
.2965
.2541
.1361
.0467
.0100
.0012
.0001
.0319
.1373
.2587
.2786
.1875
.0808
.0217
.0033
.0002
.0168
.0896
.2090
.2787
.2322
.1239
.0413
.0079
.0007
.0084
.0548
.1569
.2568
.2627
.1719
.0703
.0164
.0017
.0039
.0313
.1094
.2188
.2734
.2188
.1094
.0313
.0039
8
7
6
5
4
3
2
1
0
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
k↑
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
3
4
5
6
7
8
9
.6302
.2985
.0629
.0077
.0006
.0000
.0000
.0000
.0000
.0000
.3874
.3874
.1722
.0446
.0074
.0008
.0001
.0000
.0000
.0000
.2316
.3679
.2597
.1069
.0283
.0050
.0006
.0000
.0000
.0000
.1342
.3020
.3020
.1762
.0661
.0165
.0028
.0003
.0000
.0000
.0751
.2253
.3003
.2336
.1168
.0389
.0087
.0012
.0001
.0000
.0404
.1556
.2668
.2668
.1715
.0735
.0210
.0039
.0004
.0000
.0207
.1004
.2162
.2716
.2194
.1181
.0424
.0098
.0013
.0001
.0101
.0605
.1612
.2508
.2508
.1672
.0743
.0212
.0035
.0003
.0046
.0339
.1110
.2119
.2600
.2128
.1160
.0407
.0083
.0008
.0020
.0176
.0703
.1641
.2461
.2461
.1641
.0703
.0176
.0020
9
8
7
6
5
4
3
2
1
0
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
k↑
n=8
n=9
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 136
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
TABLE 1 – The Binomial Probability Distribution
Tabulated Values are Pr(X = k) [Values are rounded to four decimal places.]
Examples: For n = 10, π = .20, Pr(X = 1) = .2684; For n = 11, π = .75, Pr(X = 4) = .0064
n = 10
k↓
0
1
2
3
4
5
6
7
8
9
10
π = .05
.5987
.3151
.0746
.0105
.0010
.0001
.0000
.0000
.0000
.0000
.0000
.95
.10
.3487
.3874
.1937
.0574
.0112
.0015
.0001
.0000
.0000
.0000
.0000
.90
.15
.1969
.3474
.2759
.1298
.0401
.0085
.0012
.0001
.0000
.0000
.0000
.85
.20
.1074
.2684
.3020
.2013
.0881
.0264
.0055
.0008
.0001
.0000
.0000
.80
.25
.0563
.1877
.2816
.2503
.1460
.0584
.0162
.0031
.0004
.0000
.0000
.75
.30
.0282
.1211
.2335
.2668
.2001
.1029
.0368
.0090
.0014
.0001
.0000
.70
.35
.0135
.0725
.1757
.2522
.2377
.1536
.0689
.0212
.0043
.0005
.0000
.65
.40
.0060
.0403
.1209
.2150
.2508
.2007
.1115
.0425
.0106
.0016
.0001
.60
.45
.0025
.0207
.0763
.1665
.2384
.2340
.1596
.0746
.0229
.0042
.0003
.55
.50
.0010
.0098
.0439
.1172
.2051
.2461
.2051
.1172
.0439
.0098
.0010
.50
10
9
8
7
6
5
4
3
2
1
0
k↑
π = .05
.5688
.3293
.0867
.0137
.0014
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.95
.10
.3138
.3835
.2131
.0710
.0158
.0025
.0003
.0000
.0000
.0000
.0000
.0000
.90
.15
.1673
.3248
.2866
.1517
.0536
.0132
.0023
.0003
.0000
.0000
.0000
.0000
.85
.20
.0859
.2362
.2953
.2215
.1107
.0388
.0097
.0017
.0002
.0000
.0000
.0000
.80
.25
.0422
.1549
.2581
.2581
.1721
.0803
.0268
.0064
.0011
.0001
.0000
.0000
.75
.30
.0198
.0932
.1998
.2568
.2201
.1321
.0566
.0173
.0037
.0005
.0000
.0000
.70
.35
.0088
.0518
.1395
.2254
.2428
.1830
.0985
.0379
.0102
.0018
.0002
.0000
.65
.40
.0036
.0266
.0887
.1774
.2365
.2207
.1471
.0701
.0234
.0052
.0007
.0000
.60
.45
.0014
.0125
.0513
.1259
.2060
.2360
.1931
.1128
.0462
.0126
.0021
.0002
.55
.50
.0005
.0054
.0269
.0806
.1611
.2256
.2256
.1611
.0806
.0269
.0054
.0005
.50
11
10
9
8
7
6
5
4
3
2
1
0
k↑
π = .05
.5404
.3413
.0988
.0173
.0021
.0002
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.95
.10
.2824
.3766
.2301
.0852
.0213
.0038
.0005
.0000
.0000
.0000
.0000
.0000
.0000
.90
.15
.1422
.3012
.2924
.1720
.0683
.0193
.0040
.0006
.0001
.0000
.0000
.0000
.0000
.85
.20
.0687
.2062
.2835
.2362
.1329
.0532
.0155
.0033
.0005
.0001
.0000
.0000
.0000
.80
.25
.0317
.1267
.2323
.2581
.1936
.1032
.0401
.0115
.0024
.0004
.0000
.0000
.0000
.75
.30
.0138
.0712
.1678
.2397
.2311
.1585
.0792
.0291
.0078
.0015
.0002
.0000
.0000
.70
.35
.0057
.0368
.1088
.1954
.2367
.2039
.1281
.0591
.0199
.0048
.0008
.0001
.0000
.65
.40
.0022
.0174
.0639
.1419
.2128
.2270
.1766
.1009
.0420
.0125
.0025
.0003
.0000
.60
.45
.0008
.0075
.0339
.0923
.1700
.2225
.2124
.1489
.0762
.0277
.0068
.0010
.0001
.55
.50
.0002
.0029
.0161
.0537
.1208
.1934
.2256
.1934
.1208
.0537
.0161
.0029
.0002
.50
12
11
10
9
8
7
6
5
4
3
2
1
0
k↑
n = 11
k↓
0
1
2
3
4
5
6
7
8
9
10
11
n = 12
k↓
0
1
2
3
4
5
6
7
8
9
10
11
12
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 137
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
TABLE 1 – The Binomial Probability Distribution
Tabulated Values are Pr(X = k) [Values are rounded to four decimal places.]
Examples: For n = 15, π = .10, Pr(X = 6) = .0019; For n = 20, π = .85, Pr(X = 15) = .1028
n = 15
k↓
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
π = .05
.4633
.3658
.1348
.0307
.0049
.0006
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.95
.10
.2059
.3432
.2669
.1285
.0428
.0105
.0019
.0003
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.90
.15
.0874
.2312
.2856
.2184
.1156
.0449
.0132
.0030
.0005
.0001
.0000
.0000
.0000
.0000
.0000
.85
.20
.0352
.1319
.2309
.2501
.1876
.1032
.0430
.0138
.0035
.0007
.0001
.0000
.0000
.0000
.0000
.80
.25
.0134
.0668
.1559
.2252
.2252
.1651
.0917
.0393
.0131
.0034
.0007
.0001
.0000
.0000
.0000
.75
.30
.0047
.0305
.0916
.1700
.2186
.2061
.1472
.0811
.0348
.0116
.0030
.0006
.0001
.0000
.0000
.70
.35
.0016
.0126
.0476
.1110
.1792
.2123
.1906
.1319
.0710
.0298
.0096
.0024
.0004
.0001
.0000
.65
.40
.0005
.0047
.0219
.0634
.1268
.1859
.2066
.1771
.1181
.0612
.0245
.0074
.0016
.0003
.0000
.60
.45
.0001
.0016
.0090
.0318
.0780
.1404
.1914
.2013
.1647
.1048
.0515
.0191
.0052
.0010
.0001
.55
.50
.0000
.0005
.0032
.0139
.0417
.0916
.1527
.1964
.1964
.1527
.0916
.0417
.0139
.0032
.0005
.50
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
k↑
π = .05
.3585
.3774
.1887
.0596
.0133
.0022
.0003
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.95
.10
.1216
.2702
.2852
.1901
.0898
.0319
.0089
.0020
.0004
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.90
.15
.0388
.1368
.2293
.2428
.1821
.1028
.0454
.0160
.0046
.0011
.0002
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.85
.20
.0115
.0576
.1369
.2054
.2182
.1746
.1091
.0545
.0222
.0074
.0020
.0005
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.80
.25
.0032
.0211
.0669
.1339
.1897
.2023
.1686
.1124
.0609
.0271
.0099
.0030
.0008
.0002
.0000
.0000
.0000
.0000
.0000
.75
.30
.0008
.0068
.0278
.0716
.1304
.1789
.1916
.1643
.1144
.0654
.0308
.0120
.0039
.0010
.0002
.0000
.0000
.0000
.0000
.70
.35
.0002
.0020
.0100
.0323
.0738
.1272
.1712
.1844
.1614
.1158
.0686
.0336
.0136
.0045
.0012
.0003
.0000
.0000
.0000
.65
.40
.0000
.0005
.0031
.0123
.0350
.0746
.1244
.1659
.1797
.1597
.1171
.0710
.0355
.0146
.0049
.0013
.0003
.0000
.0000
.60
.45
.0000
.0001
.0008
.0040
.0139
.0365
.0746
.1221
.1623
.1771
.1593
.1185
.0727
.0366
.0150
.0049
.0013
.0002
.0000
.55
.50
.0000
.0000
.0002
.0011
.0046
.0148
.0370
.0739
.1201
.1602
.1762
.1602
.1201
.0739
.0370
.0148
.0046
.0011
.0002
.50
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
k↑
n = 20
k↓
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 138
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
n = 25
k↓
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
π = .05
.2774
.3650
.2305
.0930
.0269
.0060
.0010
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.95
.10
.0718
.1994
.2659
.2265
.1384
.0646
.0239
.0072
.0018
.0004
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.90
.15
.0172
.0759
.1607
.2174
.2110
.1564
.0920
.0441
.0175
.0058
.0016
.0004
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.85
.20
.0038
.0236
.0708
.1358
.1867
.1960
.1633
.1108
.0623
.0294
.0118
.0040
.0012
.0003
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.80
.25
.0008
.0063
.0251
.0641
.1175
.1645
.1828
.1654
.1241
.0781
.0417
.0189
.0074
.0025
.0007
.0002
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.75
.30
.0001
.0014
.0074
.0243
.0572
.1030
.1472
.1712
.1651
.1336
.0916
.0536
.0268
.0115
.0042
.0013
.0004
.0001
.0000
.0000
.0000
.0000
.0000
.70
.35
.0000
.0003
.0018
.0076
.0224
.0506
.0908
.1327
.1607
.1635
.1409
.1034
.0650
.0350
.0161
.0064
.0021
.0006
.0001
.0000
.0000
.0000
.0000
.65
.40
.0000
.0000
.0004
.0019
.0071
.0199
.0442
.0800
.1200
.1511
.1612
.1465
.1140
.0760
.0434
.0212
.0088
.0031
.0009
.0002
.0000
.0000
.0000
.60
.45
.0000
.0000
.0001
.0004
.0018
.0063
.0172
.0381
.0701
.1084
.1419
.1583
.1511
.1236
.0867
.0520
.0266
.0115
.0042
.0013
.0003
.0001
.0000
.55
.50
.0000
.0000
.0000
.0001
.0004
.0016
.0053
.0143
.0322
.0609
.0974
.1328
.1550
.1550
.1328
.0974
.0609
.0322
.0143
.0053
.0016
.0004
.0001
.50
25
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
k↑
π = .05
.2146
.3389
.2586
.1270
.0451
.0124
.0027
.0005
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.95
.10
.0424
.1413
.2277
.2361
.1771
.1023
.0474
.0180
.0058
.0016
.0004
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.90
.15
.0076
.0404
.1034
.1703
.2028
.1861
.1368
.0828
.0420
.0181
.0067
.0022
.0006
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.85
.20
.0012
.0093
.0337
.0785
.1325
.1723
.1795
.1538
.1106
.0676
.0355
.0161
.0064
.0022
.0007
.0002
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.80
.25
.0002
.0018
.0086
.0269
.0604
.1047
.1455
.1662
.1593
.1298
.0909
.0551
.0291
.0134
.0054
.0019
.0006
.0002
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.75
.30
.0000
.0003
.0018
.0072
.0208
.0464
.0829
.1219
.1501
.1573
.1416
.1103
.0749
.0444
.0231
.0106
.0042
.0015
.0005
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.70
.35
.0000
.0000
.0003
.0015
.0056
.0157
.0353
.0652
.1009
.1328
.1502
.1471
.1254
.0935
.0611
.0351
.0177
.0079
.0031
.0010
.0003
.0001
.0000
.0000
.0000
.0000
.65
.40
.0000
.0000
.0000
.0003
.0012
.0041
.0115
.0263
.0505
.0823
.1152
.1396
.1474
.1360
.1101
.0783
.0489
.0269
.0129
.0054
.0020
.0006
.0002
.0000
.0000
.0000
.60
.45
.0000
.0000
.0000
.0000
.0002
.0008
.0029
.0081
.0191
.0382
.0656
.0976
.1265
.1433
.1424
.1242
.0953
.0642
.0379
.0196
.0088
.0034
.0012
.0003
.0001
.0000
.55
.50
.0000
.0000
.0000
.0000
.0000
.0001
.0006
.0019
.0055
.0133
.0280
.0509
.0806
.1115
.1354
.1445
.1354
.1115
.0806
.0509
.0280
.0133
.0055
.0019
.0006
.0001
.50
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
k↑
n = 30
k↓
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 139
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
TABLE 1 – The Binomial Probability Distribution
n = 50
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
.0769
.2025
.2611
.2199
.1360
.0658
.0260
.0086
.0024
.0006
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0052
.0286
.0779
.1386
.1809
.1849
.1541
.1076
.0643
.0333
.0152
.0061
.0022
.0007
.0002
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0003
.0026
.0113
.0319
.0661
.1072
.1419
.1575
.1493
.1230
.0890
.0571
.0328
.0169
.0079
.0033
.0013
.0005
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0002
.0011
.0044
.0128
.0295
.0554
.0870
.1169
.1364
.1398
.1271
.1033
.0755
.0499
.0299
.0164
.0082
.0037
.0016
.0006
.0002
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0004
.0016
.0049
.0123
.0259
.0463
.0721
.0985
.1194
.1294
.1261
.1110
.0888
.0648
.0432
.0264
.0148
.0077
.0036
.0016
.0006
.0002
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0006
.0018
.0048
.0110
.0220
.0386
.0602
.0838
.1050
.1189
.1223
.1147
.0983
.0772
.0558
.0370
.0227
.0128
.0067
.0032
.0014
.0006
.0002
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0002
.0006
.0017
.0042
.0093
.0182
.0319
.0502
.0714
.0923
.1088
.1171
.1156
.1048
.0875
.0673
.0478
.0313
.0190
.0106
.0055
.0026
.0012
.0005
.0002
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0002
.0005
.0014
.0035
.0076
.0147
.0260
.0415
.0606
.0808
.0987
.1109
.1146
.1091
.0959
.0778
.0584
.0405
.0259
.0154
.0084
.0043
.0020
.0009
.0003
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0004
.0011
.0027
.0059
.0116
.0207
.0339
.0508
.0700
.0888
.1038
.1119
.1115
.1026
.0873
.0687
.0500
.0336
.0208
.0119
.0063
.0031
.0014
.0006
.0002
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0003
.0008
.0020
.0044
.0087
.0160
.0270
.0419
.0598
.0788
.0960
.1080
.1123
.1080
.0960
.0788
.0598
.0419
.0270
.0160
.0087
.0044
.0020
.0008
.0003
.0001
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
50
49
48
47
46
45
44
43
42
41
40
39
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
k↑
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 140
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
TABLE 1 – The Binomial Probability Distribution
Table Continues on Next Page!!
n = 100
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
.0059
.0312
.0812
.1396
.1781
.1800
.1500
.1060
.0649
.0349
.0167
.0072
.0028
.0010
.0003
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0003
.0016
.0059
.0159
.0339
.0596
.0889
.1148
.1304
.1319
.1199
.0988
.0743
.0513
.0327
.0193
.0106
.0054
.0026
.0012
.0005
.0002
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0003
.0011
.0031
.0075
.0153
.0276
.0444
.0640
.0838
.1001
.1098
.1111
.1041
.0908
.0739
.0563
.0402
.0270
.0171
.0103
.0058
.0031
.0016
.0008
.0004
.0002
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0002
.0006
.0015
.0034
.0069
.0128
.0216
.0335
.0481
.0638
.0789
.0909
.0981
.0993
.0946
.0849
.0720
.0577
.0439
.0316
.0217
.0141
.0088
.0052
.0029
.0016
.0008
.0004
.0002
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0003
.0006
.0014
.0030
.0057
.0100
.0165
.0254
.0365
.0493
.0626
.0749
.0847
.0906
.0918
.0883
.0806
.0701
.0580
.0458
.0344
.0248
.0170
.0112
.0070
.0042
.0024
.0013
.0007
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0002
.0006
.0012
.0024
.0044
.0076
.0124
.0190
.0277
.0380
.0496
.0613
.0720
.0804
.0856
.0868
.0840
.0776
.0685
.0579
.0468
.0362
.0268
.0191
.0130
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0002
.0004
.0009
.0017
.0032
.0055
.0090
.0140
.0207
.0290
.0388
.0494
.0601
.0698
.0774
.0821
.0834
.0811
.0755
.0674
.0577
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0001
.0003
.0006
.0012
.0022
.0038
.0063
.0100
.0151
.0217
.0297
.0391
.0491
.0591
.0682
.0754
.0799
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0002
.0004
.0008
.0014
.0025
.0043
.0069
.0106
.0157
.0222
.0301
.0391
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0001
.0001
.0002
.0005
.0009
.0016
.0027
.0045
.0071
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
100
99
98
97
96
95
94
93
92
91
90
89
88
87
86
85
84
83
82
81
80
79
78
77
76
75
74
73
72
71
70
69
68
67
66
65
64
63
62
61
k↑
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 141
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
TABLE 1 – The Binomial Probability Distribution
n = 100
k↓
π=
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0004
.0002
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0085
.0053
.0032
.0019
.0010
.0005
.0003
.0001
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0474
.0373
.0282
.0205
.0143
.0096
.0062
.0038
.0023
.0013
.0007
.0004
.0002
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0812
.0792
.0742
.0667
.0576
.0478
.0381
.0292
.0215
.0152
.0103
.0068
.0042
.0026
.0015
.0008
.0004
.0002
.0001
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0488
.0584
.0672
.0741
.0786
.0800
.0782
.0736
.0665
.0577
.0482
.0386
.0298
.0221
.0157
.0108
.0071
.0045
.0027
.0016
.0009
.0005
.0002
.0001
.0001
.0000
.0000
.0000
.0000
.0000
.0000
.0108
.0159
.0223
.0301
.0390
.0485
.0580
.0666
.0735
.0780
.0796
.0780
.0735
.0666
.0580
.0485
.0390
.0301
.0223
.0159
.0108
.0071
.0045
.0027
.0016
.0009
.0005
.0002
.0001
.0001
.0000
60
59
58
57
56
55
54
53
52
51
50
49
48
47
46
45
44
43
42
41
40
39
38
37
36
35
34
33
32
31
30
.95
.90
.85
.80
.75
.70
.65
.60
.55
.50
k↑
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 142
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
TABLE 2 – Standard Normal Curve Areas
Tabulated Values are Pr(Z ≤ z) [Values are rounded to five decimal places.]
•
•
•
From Table, Pr(Z ≤ -2.86) = .00212 so Pr(-2.86 ≤ Z ≤ 0) = ..5 - .00212 = .49788 since Pr(Z ≤ 0) = .5
Also Pr(Z ≥ -2.86) = 1 - .00212 = .99788 since Pr(Z ≥ -2.86) + Pr(Z ≤ -2.86) = 1.00000
To find k so that Pr(0 ≥ Z ≥ k) = .20000, convert to a statement in the form of Pr(Z ≤ k), work backwards
Here since Pr(0 ≥ Z ≥ k) = .20000 and Pr(Z ≤ 0) = .50000, Pr(Z ≤ k) = .30000, and so k = -0.52 is best, since Pr(Z
≤ -0.52) = .30153 is closer to desired .30000 than Pr(Z ≤ -0.53) = .29806
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
-3.00
-2.90
-2.80
-2.70
-2.60
-2.50
-2.40
-2.30
-2.20
-2.10
-2.00
-1.90
-1.80
-1.70
-1.60
-1.50
-1.40
-1.30
-1.20
-1.10
-1.00
-0.90
-0.80
-0.70
-0.60
-0.50
-0.40
-0.30
-0.20
-0.10
-0.00
.00135
.00187
.00256
.00347
.00466
.00621
.00820
.01072
.01390
.01786
.02275
.02872
.03593
.04457
.05480
.06681
.08076
.09680
.11507
.13567
.15866
.18406
.21186
.24196
.27425
.30854
.34458
.38209
.42074
.46017
.50000
.00131
.00181
.00248
.00336
.00453
.00604
.00798
.01044
.01355
.01743
.02222
.02807
.03515
.04363
.05370
.06552
.07927
.09510
.11314
.13350
.15625
.18141
.20897
.23885
.27093
.30503
.34090
.37828
.41683
.45620
.49601
.00126
.00175
.00240
.00326
.00440
.00587
.00776
.01017
.01321
.01700
.02169
.02743
.03438
.04272
.05262
.06426
.07780
.09342
.11123
.13136
.15386
.17879
.20611
.23576
.26763
.30153
.33724
.37448
.41294
.45224
.49202
.00122
.00169
.00233
.00317
.00427
.00570
.00755
.00990
.01287
.01659
.02118
.02680
.03362
.04182
.05155
.06301
.07636
.09176
.10935
.12924
.15151
.17619
.20327
.23270
.26435
.29806
.33360
.37070
.40905
.44828
.48803
.00118
.00164
.00226
.00307
.00415
.00554
.00734
.00964
.01255
.01618
.02068
.02619
.03288
.04093
.05050
.06178
.07493
.09012
.10749
.12714
.14917
.17361
.20045
.22965
.26109
.29460
.32997
.36693
.40517
.44433
.48405
.00114
.00159
.00219
.00298
.00402
.00539
.00714
.00939
.01222
.01578
.02018
.02559
.03216
.04006
.04947
.06057
.07353
.08851
.10565
.12507
.14686
.17106
.19766
.22663
.25785
.29116
.32636
.36317
.40129
.44038
.48006
.00111
.00154
.00212
.00289
.00391
.00523
.00695
.00914
.01191
.01539
.01970
.02500
.03144
.03920
.04846
.05938
.07215
.08692
.10383
.12302
.14457
.16853
.19489
.22363
.25463
.28774
.32276
.35942
.39743
.43644
.47608
.00107
.00149
.00205
.00280
.00379
.00508
.00676
.00889
.01160
.01500
.01923
.02442
.03074
.03836
.04746
.05821
.07078
.08534
.10204
.12100
.14231
.16602
.19215
.22065
.25143
.28434
.31918
.35569
.39358
.43251
.47210
.00104
.00144
.00199
.00272
.00368
.00494
.00657
.00866
.01130
.01463
.01876
.02385
.03005
.03754
.04648
.05705
.06944
.08379
.10027
.11900
.14007
.16354
.18943
.21770
.24825
.28096
.31561
.35197
.38974
.42858
.46812
.00100
.00139
.00193
.00264
.00357
.00480
.00639
.00842
.01101
.01426
.01831
.02330
.02938
.03673
.04551
.05592
.06811
.08226
.09853
.11702
.13786
.16109
.18673
.21476
.24510
.27760
.31207
.34827
.38591
.42465
.46414
z
area
-3.10
-3.20
-3.30
-3.40
-3.50
-4.00
-4.50
-5.00
.000968
.000687
.000483
.000337
.000233
.000032
.000003
.0000003
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 143
Statistics Pre-Term Program – Chapter Four
Some Special Probability Models
TABLE 2 – Standard Normal Curve Areas
Tabulated Values are Pr(Z ≤ z) [Values are rounded to five decimal places.]
•
•
From Table, Pr(Z ≤ 0.03) = .51197 so Pr(0 ≤ Z ≤ 0.03) = .01197 since Pr(Z ≤ 0) = Pr(Z ≥ 0) = .50000
Also Pr(Z ≥ 0.03) = 1 - .51197 = .48803 since Pr(Z ≥ 0.03) + Pr(Z ≤ 0.03) = 1.00000
To find k so that Pr(0 ≤ Z ≤ k) = .30000, convert to a statement in the form of Pr(Z ≤ k), work backwards
Here since Pr(0 ≤ Z ≤ k) = .30000 and Pr(Z ≤ 0) = .50000, Pr(Z ≤ k) = .80000, and so k = 0.84 is best, since Pr(Z
≤ 0.84) = .79955 is closer to desired .80000 than Pr(Z ≤ 0.85) = .80234
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
2.10
2.20
2.30
2.40
2.50
2.60
2.70
2.80
2.90
3.00
.50000
.53983
.57926
.61791
.65542
.69146
.72575
.75804
.78814
.81594
.84134
.86433
.88493
.90320
.91924
.93319
.94520
.95543
.96407
.97128
.97725
.98214
.98610
.98928
.99180
.99379
.99534
.99653
.99744
.99813
.99865
.50399
.54380
.58317
.62172
.65910
.69497
.72907
.76115
.79103
.81859
.84375
.86650
.88686
.90490
.92073
.93448
.94630
.95637
.96485
.97193
.97778
.98257
.98645
.98956
.99202
.99396
.99547
.99664
.99752
.99819
.99869
.50798
.54776
.58706
.62552
.66276
.69847
.73237
.76424
.79389
.82121
.84614
.86864
.88877
.90658
.92220
.93574
.94738
.95728
.96562
.97257
.97831
.98300
.98679
.98983
.99224
.99413
.99560
.99674
.99760
.99825
.99874
.51197
.55172
.59095
.62930
.66640
.70194
.73565
.76730
.79673
.82381
.84849
.87076
.89065
.90824
.92364
.93699
.94845
.95818
.96638
.97320
.97882
.98341
.98713
.99010
.99245
.99430
.99573
.99683
.99767
.99831
.99878
.51595
.55567
.59483
.63307
.67003
.70540
.73891
.77035
.79955
.82639
.85083
.87286
.89251
.90988
.92507
.93822
.94950
.95907
.96712
.97381
.97932
.98382
.98745
.99036
.99266
.99446
.99585
.99693
.99774
.99836
.99882
.51994
.55962
.59871
.63683
.67364
.70884
.74215
.77337
.80234
.82894
.85314
.87493
.89435
.91149
.92647
.93943
.95053
.95994
.96784
.97441
.97982
.98422
.98778
.99061
.99286
.99461
.99598
.99702
.99781
.99841
.99886
.52392
.56356
.60257
.64058
.67724
.71226
.74537
.77637
.80511
.83147
.85543
.87698
.89617
.91308
.92785
.94062
.95154
.96080
.96856
.97500
.98030
.98461
.98809
.99086
.99305
.99477
.99609
.99711
.99788
.99846
.99889
.52790
.56749
.60642
.64431
.68082
.71566
.74857
.77935
.80785
.83398
.85769
.87900
.89796
.91466
.92922
.94179
.95254
.96164
.96926
.97558
.98077
.98500
.98840
.99111
.99324
.99492
.99621
.99720
.99795
.99851
.99893
.53188
.57142
.61026
.64803
.68439
.71904
.75175
.78230
.81057
.83646
.85993
.88100
.89973
.91621
.93056
.94295
.95352
.96246
.96995
.97615
.98124
.98537
.98870
.99134
.99343
.99506
.99632
.99728
.99801
.99856
.99896
.53586
.57535
.61409
.65173
.68793
.72240
.75490
.78524
.81327
.83891
.86214
.88298
.90147
.91774
.93189
.94408
.95449
.96327
.97062
.97670
.98169
.98574
.98899
.99158
.99361
.99520
.99643
.99736
.99807
.99861
.99900
z
area
3.10
3.20
3.30
3.40
3.50
4.00
4.50
5.00
.999032
.999313
.999517
.99966
.999767
.999968
.999997
.9999997
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 144