5 Normal Probability Distributions

1
5 Normal Probability Distributions
Elementary Statistics Larson Farber
2
Section 5.1
Introduction to Normal Distributions
3
Properties of a Normal Distribution
x

• The mean, median, and mode are equal

• Bell shaped and is symmetric about the mean

• The total area that lies under the curve is one
  or 100

4
Properties of a Normal Distribution
Inflection point
Inflection point
x

• As the curve extends farther and farther away
  from the
• mean, it gets closer and closer to the x-axis but
  never touches it.

• The points at which the curvature changes are
  called inflection points. The graph curves
  downward between the inflection points and curves
  upward past the inflection points to the left and
  to the right.

5
Means and Standard Deviations
Curves with different means, same standard
deviation
12
15
18
10
11
13
14
16
17
19
20
Curves with different means, different standard
deviations
20
12
15
18
10
11
13
14
16
17
19
21
22
9
6
Empirical Rule
About 68 of the area lies within 1 standard
deviation of the mean
68
7
Determining Intervals
x
4.2
4.5
4.8
5.1
3.9
3.6
3.3
An instruction manual claims that the assembly
time for a product is normally distributed with a
mean of 4.2 hours and standard deviation 0.3
hour. Determine the interval in which 95 of
the assembly times fall.
95 of the data will fall within 2 standard
deviations of the mean.
4.2 2 (0.3) 3.6 and 4.2 2 (0.3) 4.8. 95
of the assembly times will be between 3.6 and
4.8 hrs.
8
Section 5.2
The Standard Normal Distribution
9
The Standard Score
The standard score, or z-score, represents the
number of standard deviations a random variable x
falls from the mean.
The test scores for a civil service exam are
normally distributed with a mean of 152 and a
standard deviation of 7. Find the standard
z-score for a person with a score of (a) 161
(b) 148 (c) 152
(a)
(b)
(c)
10
The Standard Normal Distribution
The standard normal distribution has a mean of 0
and a standard deviation of 1.
Using z-scores any normal distribution can be
transformed into the standard normal distribution.
z
4
3
2
1
0
1
2
3
4
11
Cumulative Areas
The total area under the curve is one.
z
0
1
2
3
1
2
3

• The cumulative area is close to 0 for z-scores
  close to 3.49.

• The cumulative area for z 0 is 0.5000.

• The cumulative area is close to 1 for z-scores
  close to 3.49.

12
Cumulative Areas
Find the cumulative area for a z-score of 1.25.
0.1056
z
0
1
2
3
1
2
3
Read down the z column on the left to z 1.25
and across to the column under .05. The value in
the cell is 0.1056, the cumulative area.
The probability that z is at most 1.25 is
0.1056.
13
Finding Probabilities
To find the probability that z is less than a
given value, read the cumulative area in the
table corresponding to that z-score.
Find P(z lt 1.45).
P (z lt 1.45) 0.0735
z
0
1
2
3
1
2
3
Read down the z-column to 1.4 and across to .05.
The cumulative area is 0.0735.
14
Finding Probabilities
To find the probability that z is greater than a
given value, subtract the cumulative area in the
table from 1.
Find P(z gt 1.24).
0.1075
0.8925
z
0
1
2
3
1
2
3
The cumulative area (area to the left) is 0.1075.
So the area to the right is 1 0.1075 0.8925.
P(z gt 1.24) 0.8925
15
Finding Probabilities
To find the probability z is between two given
values, find the cumulative areas for each and
subtract the smaller area from the larger.
Find P(1.25 lt z lt 1.17).
z
0
1
2
3
1
2
3
2. P(z lt 1.25) 0.1056
1. P(z lt 1.17) 0.8790
3. P(1.25 lt z lt 1.17) 0.8790 0.1056 0.7734
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Summary
To find the probability that z is less than a
given value, read the corresponding cumulative
area.
z
0
1
2
3
-1
-2
-3
To find the probability is greater than a given
value, subtract the cumulative area in the table
from 1.
z
0
1
2
3
-1
-2
-3
To find the probability z is between two given
values, find the cumulative areas for each and
subtract the smaller area from the larger.
z
0
1
2
3
-1
-2
-3
17
Section 5.3
Normal Distributions Finding Probabilities
18
Probabilities and Normal Distributions
If a random variable, x is normally distributed,
the probability that x will fall within an
interval is equal to the area under the curve in
the interval.
IQ scores are normally distributed with a mean of
100 and a standard deviation of 15. Find the
probability that a person selected at random will
have an IQ score less than 115.
115
100
To find the area in this interval, first find the
standard score equivalent to x 115.
19
Probabilities and Normal Distributions
Normal Distribution
Find P(x lt 115).
115
100
Standard Normal Distribution
SAME
SAME
Find P(z lt 1).
1
0
P(z lt 1) 0.8413, so P(x lt115) 0.8413
20
Application
Monthly utility bills in a certain city are
normally distributed with a mean of 100 and a
standard deviation of 12. A utility bill is
randomly selected. Find the probability it is
between 80 and 115.
P(80 lt x lt 115)
P(1.67 lt z lt 1.25)
0.8944 0.0475 0.8469
The probability a utility bill is between 80 and
115 is 0.8469.
21
Section 5.4
Normal Distributions Finding Values
22
From Areas to z-Scores
Find the z-score corresponding to a cumulative
area of 0.9803.
z 2.06 corresponds roughly to the 98th
percentile.
0.9803
4
3
2
1
0
1
2
3
4
z
Locate 0.9803 in the area portion of the table.
Read the values at the beginning of the
corresponding row and at the top of the column.
The z-score is 2.06.
23
Finding z-Scores from Areas
Find the z-score corresponding to the 90th
percentile.
.90
z
0
The closest table area is .8997. The row heading
is 1.2 and column heading is .08. This
corresponds to z 1.28.
A z-score of 1.28 corresponds to the 90th
percentile.
24
Finding z-Scores from Areas
Find the z-score with an area of .60 falling to
its right.
.40
.60
z
0
z
With .60 to the right, cumulative area is .40.
The closest area is .4013. The row heading is
0.2 and column heading is .05. The z-score is
0.25.
A z-score of 0.25 has an area of .60 to its
right. It also corresponds to the 40th percentile
25
Finding z-Scores from Areas
Find the z-score such that 45 of the area under
the curve falls between z and z.
.275
.275
.45
z
z
The area remaining in the tails is .55. Half this
area is in each tail, so since .55/2 .275 is
the cumulative area for the negative z value and
.275 .45 .725 is the cumulative area for the
positive z. The closest table area is .2743 and
the z-score is 0.60. The positive z score is 0.60.
26
From z-Scores to Raw Scores
To find the data value, x when given a standard
score, z
The test scores for a civil service exam are
normally distributed with a mean of 152 and a
standard deviation of 7. Find the test score for
a person with a standard score of (a) 2.33
(b) 1.75 (c) 0
(a) x 152 (2.33)(7) 168.31
(b) x 152 (1.75)(7) 139.75
(c) x 152 (0)(7) 152
27
Finding Percentiles or Cut-off Values
Monthly utility bills in a certain city are
normally distributed with a mean of 100 and a
standard deviation of 12. What is the smallest
utility bill that can be in the top 10 of the
bills?
115.36 is the smallest value for the top 10.
90
10
z
Find the cumulative area in the table that is
closest to 0.9000 (the 90th percentile.) The area
0.8997 corresponds to a z-score of 1.28.
To find the corresponding x-value, use
x 100 1.28(12) 115.36.
28
Section 5.5
The Central Limit Theorem
29
Sampling Distributions
A sampling distribution is the probability
distribution of a sample statistic that is formed
when samples of size n are repeatedly taken from
a population. If the sample statistic is the
sample mean, then the distribution is the
sampling distribution of sample means.
Sample
Sample
Sample
Sample
Sample
Sample
The sampling distribution consists of the values
of the sample means,

30
The Central Limit Theorem
If a sample n ? 30 is taken from a population
with any type distribution that has a mean and
standard deviation
x
the sample means will have a normal distribution
and standard deviation
31
The Central Limit Theorem
If a sample of any size is taken from a
population with a normal distribution with mean
and standard deviation
x
the distribution of means of sample size n, will
be normal with a mean standard deviation
32
Application
The mean height of American men (ages 20-29) is
inches. Random samples of 60 such men
are selected. Find the mean and standard
deviation (standard error) of the sampling
distribution.
mean
69.2
Distribution of means of sample size 60, will be
normal.
Standard deviation
33
Interpreting the Central Limit Theorem
The mean height of American men (ages 20-29) is
69.2. If a random sample of 60 men in this
age group is selected, what is the probability
the mean height for the sample is greater than
70? Assume the standard deviation is 2.9.
Since n gt 30 the sampling distribution of
will be normal
mean
standard deviation
Find the z-score for a sample mean of 70
34
Interpreting the Central Limit Theorem
2.14
z
There is a 0.0162 probability that a sample of 60
men will have a mean height greater than 70.
35
Application Central Limit Theorem
During a certain week the mean price of gasoline
in California was 1.164 per gallon. What is the
probability that the mean price for the sample of
38 gas stations in California is between 1.169
and 1.179? Assume the standard deviation
0.049.
Since n gt 30 the sampling distribution of
will be normal
mean
standard deviation
Calculate the standard z-score for sample values
of 1.169 and 1.179.
36
Application Central Limit Theorem
P( 0.63 lt z lt 1.90) 0.9713 0.7357 0.2356
z
.63
1.90
The probability is 0.2356 that the mean for the
sample is between 1.169 and 1.179.
37
Section 5.6
Normal Approximation to the Binomial
38
Binomial Distribution Characteristics
There are a fixed number of independent trials.
(n) Each trial has 2 outcomes, Success or
Failure. The probability of success on a single
trial is p and the probability of failure is q.
p q 1 We can find the probability of
exactly x successes out of n trials. Where x 0
or 1 or 2 n. x is a discrete random
variable representing a count of the number of
successes in n trials.
39
Application
34 of Americans have type A blood. If 500
Americans are sampled at random, what is the
probability at least 300 have type A blood?
Using techniques of Chapter 4 you could calculate
the probability that exactly 300, exactly 301
exactly 500 Americans have A blood type and add
the probabilities.
Oryou could use the normal curve probabilities
to approximate the binomial probabilities.
If np ? 5 and nq ? 5, the binomial random
variable x is approximately normally distributed
with mean
40
Why Do We Require np ? 5 and nq ? 5?
n 5 p 0.25, q .75 np 1.25 nq 3.75
0
1
2
3
4
5
n 20 p 0.25 np 5 nq 15
0
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3
1
4
1
5
1
6
1
7
1
8
1
9
2
0
n 50 p 0.25 np 12.5 nq 37.5
0
10
20
30
40
50
41
Binomial Probabilities
The binomial distribution is discrete with a
probability histogram graph. The probability that
a specific value of x will occur is equal to the
area of the rectangle with midpoint at x.
If n 50 and p 0.25 find
Add the areas of the rectangles with midpoints at
x 14, x 15,
x 16.
0.111 0.089 0.065 0.265
0.111
0.089
0.065
14
15
16
42
Correction for Continuity
Use the normal approximation to the binomial to
find
.
14
15
16
Values for the binomial random variable x are 14,
15 and 16.
43
Correction for Continuity
Use the normal approximation to the binomial to
find
.
14
15
16
The interval of values under the normal curve is
To ensure the boundaries of each rectangle are
included in the interval, subtract 0.5 from a
left-hand boundary and add 0.5 to a right-hand
boundary.
44
Normal Approximation to the Binomial
Use the normal approximation to the binomial to
find
.
Find the mean and standard deviation using
binomial distribution formulas.
Adjust the endpoints to correct for continuity P
.
Convert each endpoint to a standard score.
45
Application
A survey of Internet users found that 75 favored
government regulations of junk e-mail. If 200
Internet users are randomly selected, find the
probability that fewer than 140 are in favor of
government regulation.
Since np 150 ? 5 and nq 50 ? 5 use the
normal approximation to the binomial.
The binomial phrase of fewer than 140 means 0,
1, 2, 3139.
Use the correction for continuity to translate to
the continuous variable in the interval
. Find P( x lt 139.5).
46
Application
A survey of Internet users found that 75 favored
government regulations of junk e-mail. If 200
Internet users are randomly selected, find the
probability that fewer than 140 are in favor of
government regulation.
Use the correction for continuity P(x lt 139.5).
P( z lt -1.71) 0.0436
The probability that fewer than 140 are in favor
of government regulation is approximately 0.0436.