StatsChap7

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Chapter
7
The Normal
Probability
Distribution
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Chap 2
2
Jury Selection Problem
There are 60 women and 40 men waiting
outside the courtroom in the jury pool.
Assuming the probability of being
selected for the jury is the same for
each person, what is the probability
that the 12 person jury selected will
consist of 7 men and 5 women?
3
Section
7.1
Properties
of the
Normal
Distribution
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
A probability density function (pdf) is an
equation used to compute probabilities of
continuous random variables.
The total area under the graph of the equation over all
possible values of the random variable must equal 1.
The area under the graph of the density function over an
interval represents the probability of observing a value
of the random variable in that interval.
5-5
“Probability Density Function” is the equation of the blue
curve. This equation gives the “y” values of the curve as
“x” moves from left to right.
The “area under the curve” is the area between the curve
(blue line) and the horizontal.
7-6
Relative frequency histograms that are
symmetric /bell-shaped are said to have the
shape of a normal curve.
7-7
If a continuous random variable is “normally
distributed”, or has a “normal probability
distribution”, then a rel freq histogram of that
variable has the shape of a “normal” or symmetric
bell-shaped curve
7-8
Properties of the Normal Curve
1. It is symmetric about its mean, μ.
2. Mean = median = mode. The curve has a
single maximum point at x = μ.
3. It has inflection points at μ ± σ
4. The area under the curve is 1.
5. The area under the curve to the right of μ,
as well as to the left of μ, equals 1/2.
5-9
7-10
6. As “x” increases without bound (gets further
from the mean), the graph approaches, but
never reaches, the horizontal axis.
7. The Empirical Rule (68-95-99.7):
Approximately 68% of the area under the
normal curve lies between μ ± 1 σ; approx.
95% of the area lies between μ ± 2σ; and
approx. 99.7% of the area lies between
μ ± 3σ .
5-11
7-12
A Normal Continuous Random Variable
The data on the next slide represent the heights
(in) of a random sample of 50 two-year old males.
(a) Draw a histogram of the data using first class
boundaries of 31.5 and 32.5 in.
(b) Do you think that the variable:
“x” = height of 2-year old males
is normally distributed?
7-13
HEIGHT (IN) OF TWO-YR OLD MALES
36.0
34.7
34.4
33.2
35.1
38.3
37.2
36.2
33.4
35.7
36.1
35.2
33.6
39.3
34.8
37.4
37.9
35.2
34.4
39.8
36.0
38.2
39.3
35.6
36.7
37.0
34.6
31.5
34.0
33.0
36.0
37.2
38.4
37.7
36.9
36.8
36.0
34.8
35.4
36.9
35.1
33.5
35.7
35.7
36.8
34.0
37.0
35.0
35.7
38.9
7-14
7-15
In the next slide, we have a normal density
curve superimposed over the histogram.
How does the area of the rectangle
corresponding to a height between 34.5 and
35.5 inches relate to the area under the
curve between these two heights?
7-16
7-17
Area under a Normal Curve
Suppose that a random variable “x” is normally
distributed with mean μ and standard deviation σ.
The area under the normal curve for any interval of
values of “x” (say, 34.5 – 35.5 in) represents either:
1. the proportion of the population with that interval
(range) of values or:
2. the probability that a randomly selected individual
from the population will have that range of values.
5-18
EXAMPLE: Area Under a Normal Curve
Suppose, the weights of adult African giraffes
are normally distributed with mean μ = 2200
pounds and standard deviation σ = 200 pounds.
(a) Shade the area under the normal curve to
the left of x = 2100 pounds.
(b) Suppose that the area under the normal
curve to the left of x = 2100 pounds is 0.3085.
Provide two (Proportion/Probability)
interpretations of this result.
7-19
EXAMPLE
Area Under a Normal Curve
Adult Giraffe: μ = 2200 pounds and σ = 200 pounds
1) The proportion of giraffes whose weight is less than 2100
pounds is 0.3085 or 30.85% of the giraffe population.
2) The probability that a randomly selected giraffe weighs
less than 2100 pounds is 0.3085.
7-20
Section
7.2
Applications of
the Normal
Distribution
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Standardizing a Normal Random Variable
Suppose that the random variable X is normally
distributed with mean μ and standard deviation
σ. Then the random variable
Z
X

is normally distributed with mean μ = 0 and
standard deviation σ = 1.The random variable Z
has a “standard” normal distribution.
5-22
“Standard” Normal Distribution
7-23
IQ scores can be modeled by a normal
distribution with μ = 100 and σ = 15.
An IQ score of 120, is 1.33 standard
deviations above the mean.
z
x

120  100

 1.33
15
7-24
Normal Probability Accuracy
For these probabilities, we require
4-decimal accuracy, so set your
calculator to:
Mode: Float: 4
25
Table IV: the area under the standard normal
curve to the left of z = 1.33 is 0.9082.
7-26
Find the area to the right of z = 1.33.
7-27
Areas Under the Standard Normal Curve
Empirical Rule: 68-95-99.7
7-28
Find the area under the standard normal curve to the
left of z = – 0.38.
Area to the left of z = – 0.38 is 0.3520.
7-29
Find the area under the standard normal curve to
the right of z = 1.25
Area right of 1.25 = 1 – area left of 1.25
= 1 – 0.8944 = 0.1056
7-30
Find the area under the standard normal curve
between z = –1.02 and z = 2.94
Area between z = –1.02 and 2.94
= (Area left of z = 2.94) – (area left of z = –1.02)
= 0.9984 – 0.1539
= 0.8445
7-31
TI-84
2:Distr: normcdf… (lo, up, mn, stddev)
(using z-score): normcdf (-1,1) = 0.6827
normcdf (-3.49,-0.38) = 0.3517
normcdf (-1E99,-0.38) = 0.3520
TI-84
2:Distr: normcdf… (lo, up, mn, stddev)
If you wanted to know the probability of
getting 120 or less on an IQ test whose mean
is 100 and stddev is 15…

(using x-score) normcdf (0,120,100,15) = 0.9088

(using z-score): normcdf (-1E99,1.333) = 0.9087
(using lo bd of z= -3.49 gives 0.9085)
x
invNorm (area/prob , mean, stddev)
Find the raw score/z-score that corresponds to
P85 on the standard IQ Test (100;15)

(using x-score) InvNorm (0.85 ,100,15)
= 115.55

(using z-score) InvNorm (0.85) = 1.0364
TI-84 sample problems…

normcdf (-1.5,1.25) → 0.8275

normcdf (0,133,100,15) → 0.9861
(89,133,100,15) → 0.7544
invNorm (0.9861) → 2.2001
 invNorm (0.9861,100,15) → 133.0015


normcdf (-1E99,1.25) → 0.8944
Finding the Value of a Normal Variable
The combined (verbal + quantitative) score on the GRE
(Graduate Record Exam (GRE ~ to get accepted into
graduate school) is normally distributed with
mean=1049 and stddev =189
What is the raw test score for a student whose rank is
at the 85th percentile?
7-36
Value of a Normal Random Variable
The z-score that corresponds to the 85th percentile
is the z-score which
has .8500 area
to its left.
TI-84 or Table V : InvNorm (0.8500) = 1.0364
So, this z-score is 1.0364.
x = µ + zσ
= 1049 + 1.0364(189)
= 1244.88 or 1245
A person who scores 1245 on the GRE would rank in
the 85th percentile, or the prob a person scores 1245 on
the GRE is 0.850
7-37
Value of a Normal Random Variable
Karlos, Inc. manufactures construction-grade steel rebar
rods. Their length is normally distributed with a mean of
100 cm (1 m) and a standard deviation of 0.45 cm.
Suppose Quality Control wants to accept 90% of all rods
manufactured, but rebuild all rods outside that tolerance.
Determine the min/max length of these rods that make up
the middle 90% of all rods manufactured.
7-38
Value of a Normal Random Variable
Area = 0.05
Area = 0.05
InvNorm(0.05) = - 1.645
z1 = –1.645 and z2 = 1.645
x1 = µ + z1σ
= 100 + (–1.645)(0.45)
= 99.2598 cm
OR
InvNorm(0.05,100,0.45)
= 99.2598 = 99.3 cm
The steel rods that make up the middle 90% would
have lengths between 99.3 cm and 100.7 cm.
7-39
Section
7.3
Assessing
Normality
(Skip)
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Section
7.4
The Normal
Approximation
to the Binomial
Probability
Distribution
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Criteria for a Binomial Probability Experiment
An experiment is said to be a binomial experiment if:
1. The experiment is performed “n” independent times.
Each repetition is called a trial. Independence means that
the outcome of one trial will not affect the outcome of the
other trials.
2. For each trial, there are only two possible disjoint
outcomes: Success / Failure (T/F; Yes/No, H/T, etc.)
3. The probability of success, “p” and the prob of failure
“q”, are the same for each trial of the experiment.
Note: p + q =1
and
q = 1- p.
5-42
For a fixed prob of success “p”, as the number
of trials “n” in a binomial experiment increases,
the probability distribution of “x” becomes
more nearly symmetric and bell-shaped.
As a rule of thumb, if the variance = npq > 10,
then the probability distribution will be approx
normal (bell-shaped) and we can use “z” scores
to predict probability of experiment outcomes.
7-43
The Normal Approximation to the Binomial
Probability Distribution
If npq ≥ 10, (stddev = npq ≥ 3.2) , the
binomial random variable “x” is approximately
normally distributed with :
Mean: μX = np
Std Dev:  X 
np 1 p .
Note: “q” = (1- p)
5-44
P(X = 18) ≈ P(17.5 < X < 18.5)
7-45
P(X < 18) ≈ P(X < 18.5)
7-46
Exact Probability Using Binomial: P(X ≤ a)
Approximate Probability
Using Normal:
P(X ≤ a + 0.5)
7-47
Exact Probability Using Binomial: P(X ≥ a)
Approximate Probability
Using Normal:
P(X ≥ a – 0.5)
7-48
Exact Probability
Using Binomial:
P(a ≤ X ≤ b)
Approximate Probability
Using Normal:
P(a=0.5 ≤ X ≤ b+0.5)
7-49
Using the Binomial Probability Distribution Function
35% of all car-owning households have three or more cars.
(a) In a sample of 400 car-owning households, what is the
probability that fewer than 150 have three or more cars?
np(1  p)  400(0.35)(1  0.35)
 91  10
 X  400(0.35)
 140
 X  400(0.35)(1  0.35)
 9.54
P(x<150) ≈ P(x<150.5)
x
150.5  140
P( z 

)

9.54
P ( z  1.006)
 0.8428
7-50
Using the Binomial Probability Distribution Function
35% of all car-owning households have three or more
cars.
(b) In a sample of 400 car-owning households, what is the
probability that at least 160 have three or more cars?
P(X  160)  P(X  159.5)
159.5  140 

 P Z 


9.54
 0.0207
7-51
Chap 2
52
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