CHAPTER 8
Section 8.4
1.
How to list the probabilities for a binomial experiment.
For a binomial experiment there are n independent ‘trials’, where n is a fixed number,
and the probability of ‘success’ remains the same from one trial to the next, and this
probability is denoted by p. The variable of interest is X: number of successes in the n
trials. The possible values of X are 0, 1, 2, …, n. In the example below, n=10 and p=0.25
Step 1: Create a column with the possible values of X. From the menu, select
Calc > Make Patterned Data > Simple set of numbers
and fill in the dialog boxes as shown below.
The worksheet now contains the column C1, that we can name X, with the possible
values of the variable from 0 to 10.
Step 2: Calculate the probability for each one of the possible values of X. From the
menu, select
Calc > Probability Distributions > Binomial
Make sure to select option ● Probability. In the Number of trials dialog box, type 10
(this is the value of n). In the Event Probability dialog box enter 0.25 (this is the value of
p). We also need to indicate that column C1 contains the values of X and the
probabilities for each one of these values are to be stored in column C2.
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After clicking on OK, the worksheet will contain the values of X and their corresponding
probabilities (P(X=k)). Note that column C2 has been named P(x).
(Optional) You can use Graph > Probability Distribution Plot > View Single > OK to
view the probability histogram. Select Binomial, enter 10 for the Number of trials, and
enter .25 for the Event probability.
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Clicking OK produces the following graph.
Distribution Plot
Binomial, n=10, p=0.25
0.30
0.25
Probability
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
X
2.
How to find exact and cumulative probabilities for a specific value of x,
given n and p. (See Example 8.10)
Given that the probability that any birth is a girl is 0.488, we can use Minitab to find the
probability of exactly seven girls in ten births, P(X=7). A second option would be to
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produce the complete table of values along with their respective probabilities and locate
the probability of interest.
Option 1. Calculating the probability for an specific value of x
From the menu, select Calc > Probability Distributions > Binomial.
Type 10 in the Number of trials: dialog box and 0.488 in the Event Probability: dialog
box. Since we want to calculate P(X=7), select Probability and type the number 7 in the
Input constant: dialog box.
Clicking OK produces the following output in the Session Window.
Probability Density Function
Binomial with n = 10 and p = 0.488
x
7
P( X = x )
0.106152
To find the probability of at most 7 girls in 10 births, we need to calculate the cumulative
probability P(X  7). To obtain the cumulative probability, select from the menu
Calc > Probability Distributions > Binomial
Indicate the Number of trials: and the Event Probability: in the spaces provided, check
Cumulative probability and type 7 in the Input constant: dialog box.
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By leaving the Optional storage: dialog box empty the results will appear in the Session
Window. (Note: If you want to store this probability then you enter a constant (e.g. K1)
in the Optional storage: dialog box. Minitab uses the letter K for constants. You must
select Data>Display Data from the menu, and then specify the constant in the dialog box
to produce the result.)
Cumulative Distribution Function
Binomial with n = 10 and p = 0.488
x
7
P( X <= x )
0.953257
(Alternatively) Use Graph > Probability Distribution Plot > View Probability > OK
and select Binomial and fill-in the dialog boxes as shown below.
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Select Shaded Area > Left Tail, X Value and enter 7 in the dialog box.
Clicking OK produces the probability (.953) along with the shade region (X < 7). Note
that Minitab has rounded the result to 3 digits.
Distribution Plot
Binomial, n=10, p=0.488
0.25
0.953
Probability
0.20
0.15
0.10
0.05
0.00
X
7
9
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To calculate the probability of more than 7 girls in 10 births, use the previous result and
subtract it from 1:
P(X>7) = 1 - P(X  7) = 1 - 0.953257 = 0.046743
To calculate the probability of at least 7 girls, observe that P(X  7) = 1-P(X  6) and
again select the Cumulative probability but enter the number 6 in the Input constant:
dialog box. The output in the Session Window gives the value of P(X  6):
Cumulative Distribution Function
Binomial with n = 10 and p = 0.488
x
6
P( X <= x )
0.847104
Therefore P(X  7) = 1 - P(X  6) = 1 - 0.847104 = 0.152896.
(Alternatively) Use Graph > Probability Distribution Plot > View Probability > OK
and select Binomial with the number of trials equal to 10 and the event probability equal
to .488. Select Shaded Area > Right Tail, X Value and enter 7 in the dialog box. You
should get the following.
Distribution Plot
Binomial, n=10, p=0.488
0.25
Probability
0.20
0.15
0.10
0.05
0.00
0.153
0
X
7
Option 2. To produce the complete table of values along with their respective
probabilities, first select Calc>Make Patterned Data> Simple Set of numbers to
produce values of X (0 to 10 in steps of 1) in a column, and then select Calc>Probability
Distributions >Binomial to calculate the probabilities for each one of the values of X.
The output will appear in the Session Window when the Optional storage: dialog box is
left empty. Recall that entering a column name (C2) in the Optional storage: dialog box
will produce results in the worksheet.
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Probability Density Function
Binomial with n = 10 and p = 0.488
x
0
1
2
3
4
5
6
7
8
9
10
P( X = x )
0.001238
0.011799
0.050607
0.128626
0.214545
0.245386
0.194903
0.106152
0.037941
0.008036
0.000766
So the probability of 7 girls in 10 births is P(X=7) = 0.106152.
To find the cumulative probabilities, P(X  x), for x=0, 1, …, 10, we need to select
Cumulative probability in the Binomial Distribution window. If the Optional storage:
dialog box is left empty then the results are printed in the Session Window.
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Cumulative Distribution Function
Binomial with n = 10 and p = 0.488
x
0
1
2
3
4
5
6
7
8
9
10
P( X <= x )
0.00124
0.01304
0.06364
0.19227
0.40682
0.65220
0.84710
0.95326
0.99120
0.99923
1.00000
The probability of at most 7 girls in 10 births is
P(X  7) = 0.95326
The probability of more than 7 girls in 10 births is
P(X>7) = 1 - P(X  7) = 1 - 0.9533 = 0.04674
The probability of at least 7 girls in 10 births is
P(X  7) = 1- P(X  6) = 1 - 0.84710 = 0.15290
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Section 8.5 Finding probabilities for a uniform distribution.
In Example 8.13, the variable X is ‘waiting time until the next bus arrives.’ The random
variable X can be any value between 0 and 10 minutes (buses come every ten minutes, so
10 minutes is the longest time one can wait at the bus stop).
X is a continuous random variable with a uniform distribution on the interval from 0 to
10 minutes; the density ‘curve’ is shown in Figure 8.2. The height of the ‘curve’ is 0.1 for
all values of X.
Since X is continuous it can take on any value in the interval [0, 10]. To show the height
of the density ‘curve’ for any value of X in the interval, select from the menu
Calc > Probability Distributions > Uniform. Then select Probability density, indicate
the minimum and maximum values of the variable (Lower endpoint: 0, Upper endpoint:
10), and enter a number in the Input constant: dialog box.
Example 8.13 continued. To find the value of P(5 ≤ X ≤ 7), i.e. the probability that the
waiting time is between 5 and 7 minutes, first we need to calculate the cumulative
probabilities for P(X  5) and P(X  7). To obtain the uniform probability window, from
the menu select
Calc > Probability Distributions > Uniform
Then select Cumulative probability and fill in the dialog boxes as shown below. If the
Optional storage: dialog box is left empty then the results appear in the Session Window.
Repeat the previous steps but enter 7 in the Input constant: dialog box. Note that the
numbers 5 and 7 could have been typed in a column and the cumulative probabilities for
these values could have been found using the Input column: dialog box option.
The two outputs will be
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x
5.0
P( X <= x)
0.5000
x
7.0
P( X <= x)
0.7000
Therefore the P(5  X  7) = P(X  7) - P(X  5) = 0.7 - 0.5 = 0.2, the same answer found
using the formula for the area of the rectangle (base x height = 2 x 0.1 = 0.2).
(Alternatively) Use Graph > Probability Distribution Plot > View Probability > OK
and select Uniform and fill-in the dialog boxes as shown below.
Select Shaded Area > Middle and fill-in the dialog boxes as shown below.
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Clicking OK will produce the result that is shown in Figure 8.3.
Distribution Plot
Uniform, Lower=0, Upper=10
0.2
0.10
Density
0.08
0.06
0.04
0.02
0.00
0
5
X
7
10
Section 8.6
1.
Finding standard normal probabilities
Assume that Z is a standard normal random variable. To find P(Z  1.31)
(area under the curve up to the point 1.31), as in Example 8.15, from the menu select
Calc > Probability Distributions > Normal
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Indicate the mean and standard deviation corresponding to the standard normal
distribution, click on Cumulative probability because we are calculating P(Z  1.31),
select Input constant: and enter 1.31 in the dialog box.
The answer will appear in the Session Window:
Cumulative Distribution Function
Normal with mean = 0 and standard deviation = 1
x
1.31
P( X <= x )
0.904902
Note. Because the normal is a continuous distribution P(Z  1.31) = P(Z<1.31); i.e. the
inclusion of the point 1.31 does not make a difference. This is not true in general for
discrete distributions.
To find P(Z>1.31), remember that P(Z>1.31) = 1 - P(Z  31)=1 - 0.904902 = 0.095098.
Minitab does not provide the area under the curve to the right of a point. Therefore, we
need to find the area under the curve to the left of the point and then subtract it from 1.
(Alternatively) Use Graph > Probability Distribution Plot > View Probability > OK
and select Normal with Mean 0 and Standard deviation 1.
Select Shaded Area > Left Tail, X value and enter 1.31 in the dialog box.
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The result appears below.
Distribution Plot
Normal, Mean=0, StDev=1
0.4
Density
0.3
0.905
0.2
0.1
0.0
0
X
1.31
To find P(Z>1.31). Follow the previous instructions that produced the above graph but
select Right Tail. You should get the following graph which is similar to Figure 8.5 of
the textbook.
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Distribution Plot
Normal, Mean=0, StDev=1
0.4
Density
0.3
0.2
0.1
0.0951
0.0
0
X
1.31
Note. To find the probability that Z is between two values follow the previous
instructions but select Middle. For example, to find P(-2.59 < Z < 1.31) replicate the
Minitab window below.
Clicking OK produces the result below which is similar to Figure 8.6.
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Distribution Plot
Normal, Mean=0, StDev=1
0.4
0.900
Density
0.3
0.2
0.1
0.0
2.
-2.59
0
X
1.31
Finding probabilities for any normal distribution.
Areas under any normal curve, not only the standard normal, can be found using Minitab.
In Example 8.14, it is assumed that college women’s heights (X) follow a normal curve
with mean μ = 65 inches and standard deviation σ = 2.7 inches.
To find the proportion of college women that are taller than 68 inches, we need to
calculate P(X>68). We need to first find the cumulative probability of 68, i.e. P(X<68).
From the menu, select
Calc > Probability Distributions > Normal
Select Cumulative probability, indicate the mean 65 and the standard deviation 2.7 of the
normal distribution, and enter the value 68 in the Input constant: dialog box.
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The answer will appear in the Session Window:
Cumulative Distribution Function
Normal with mean = 65 and standard deviation = 2.7
x
68
P( X <= x )
0.866740
Using the output, P(X>68) = 1 - 0.86674 = 0.13326. So, about 13% of college women are
taller than 68 inches.
The area under the curve between x=62 and x=68 represents the proportion of college
women that are between 62 and 68 inches tall. To find that area, P(62  X  68), we need
first to calculate the area to the left of x=62. From the menu, select Calc > Probability
Distributions > Normal and enter the mean 65 and standard deviation 2.7 of the
distribution, select Cumulative probability and in the dialog box for the Input constant:
option enter 62 to find P(X  62).
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The answer will appear in the Session Window:
Cumulative Distribution Function
Normal with mean = 65 and standard deviation = 2.7
x
62
P( X <= x )
0.133260
From a previous calculation we know P(X  68) = 0.86674. Hence, the area under the
curve between 62 and 68 can be calculated by subtracting the area under the curve to the
left of 62 from the area under the curve to the left of 68, i.e. P(62  X  68) = P(X  68) P(X  62) = 0.86674 - 0.13326 = 0.73348.
(Alternatively) Use Graph > Probability Distribution Plot > View Probability > OK
and select Normal with Mean 65 and Standard deviation 2.7. Select Shaded
Area>Middle, X value 1 is 62 and the X value 2 is 68. Clicking OK produces the desired
P(62 < X < 68) computation.
Distribution Plot
Normal, Mean=65, StDev=2.7
0.16
0.733
0.14
0.12
Density
0.10
0.08
0.06
0.04
0.02
0.00
62
65
X
68
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3.
Finding percentiles for a normal distribution
In Example 8.17, it is assumed that the blood pressures of men, in the age group 18-29
years, can be described with a normal curve with mean μ =120 and standard deviation σ
=10. We would like to know what is the value x such that 75% of the men in that age
group have a blood pressure that is at most x. In other words, we need to find x such that
P(Blood pressure  x)=0.75. To find the 75th percentile, from the menu, select
Calc>Probability Distributions> Normal
Type the mean 120 and standard deviation 10 of the distribution, select the option Inverse
cumulative probability and type the value 0.75 (because we are working with the 75th
percentile) in the Input constant: dialog box.
The answer will appear in the Session window.
Inverse Cumulative Distribution Function
Normal with mean = 120 and standard deviation = 10
P( X <= x )
0.75
x
126.745
Approximately 75% of the men, in the age group 18-29 years, have a systolic blood
pressure below 127.
(Alternatively) Use Graph > Probability Distribution Plot > View Probability > OK
and select Normal with Mean 120 and Standard deviation 10.
Select Shaded Area, Probability and Left Tail with Probability: equal to .75.
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Clicking OK produces the result rounded to127.
Distribution Plot
Normal, Mean=120, StDev=10
0.04
Density
0.03
0.02
0.75
0.01
0.00
120
X
127
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