Transcript prob_stats_ch05

5 Normal Probability Distributions
Elementary Statistics
Larson
Farber
Section 5.1
Introduction to
Normal Distributions
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%
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.
Empirical Rule - Review
68%
About 68% of the area
lies within 1 standard
deviation of the mean
About 95% of the area
lies within 2 standard
deviations
About 99.7% of the area lies within
3 standard deviations of the mean
Determining Intervals
x
3.3 3.6 3.9 4.2
4.5 4.8 5.1
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.
The Standard Normal Distribution
A normal distribution with a mean of 0 and a standard
deviation of 1 is called the standard normal distribution.
Using z-scores any normal distribution can be
transformed into the standard normal distribution.
–4 –3 –2 –1
0 1
2 3
4
z
The Standard Score - Review
The standard score, or z-score, represents the number of
standard deviations a random variable x falls from the
mean.
The Standard Score
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)
Cumulative Areas
The
total
area
under
the curve
is one.
–3 –2 –1 0 1 2 3
z
-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.
Cumulative Areas
Find the cumulative area for a z-score of –1.25.
0.1056
–3 –2 –1 0 1 2 3
z
On pages A16-A17, read down the z column on the left to
z = –1.2 and across to the column under .05 or
The probability that z is at most –1.25 is 0.1056.
Cumulative Areas
Find the cumulative area for a z-score of –1.25.
0.1056
–3 –2 –1 0 1 2 3
z
Use a graphing calculator normalcdf (-100,-1.25). The value is
0.1056, the cumulative area.
The probability that z is at most –1.25 is 0.1056.
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 < –1.45).
P (z < –1.45) = 0.0735
–3 –2 –1
0 1
2 3
Read down the z-column to –1.4 and across to .05 or use
normalcdf (-100,-1.45). The cumulative area is 0.0735.
z
Finding Probabilities
To find the probability that z is greater than a given
value, subtract the cumulative area in the table from 1 or
Use normalcdf (-1.24,100)
Find P(z > –1.24).
0.1075
0.8925
z
–3 –2 –1 0 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 > –1.24) = 0.8925
Finding Probabilities
To find the probability z is between two given values, find the
areas for each and subtract the smaller area from the larger or
Use normalcdf(-1.25,1.17).
Find P(–1.25 < z < 1.17).
–3 –2 –1 0 1 2
1. P(z < 1.17) = 0.8790
3
z
2. P(z < –1.25) = 0.1056
3. P(–1.25 < z < 1.17) = 0.8790 – 0.1056 = 0.7734
Summary
To find the probability that z is less
than a given value, read the
corresponding cumulative area.
-3 -2 -1 0 1 2 3
z
To find the probability is greater
than a given value, subtract the
cumulative area in the table from 1.
-3 -2 -1 0 1 2 3
z
To find the probability z is
between two given values, find the
cumulative areas for each and
subtract the smaller area from the
larger.
-3 -2 -1 0 1 2 3
z
Section 5.2
Normal Distributions
Finding Probabilities
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.
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.
100 115
To find the area in this interval, first find the standard
score equivalent to x = 115.
115-100 = 1
15
Probabilities and Normal Distributions
Normal Distribution
Standard Normal
Distribution
100 115
Find P(z < 1).
0 1
P(z < 1) = 0.8413, so P(x <115) = 0.8413
SAME
SAME
Find P(x < 115).
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.
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. Using pages A16-A17 or a graphing calculator.
Normal Distribution
P(80 < x < 115) = P(–1.67 < z < 1.25) =normalcdf(-1.67,1.25)
= 0.8469
The probability a utility bill is
between $80 and $115 is 0.8469
or 84.7%.
Section 5.3
Normal Distributions
Finding Values
From Areas to z-Scores
Find the z-score corresponding to a cumulative area of 0.9803.
0.9803
–4 –3 –2 –1 0
1
2
3
4
z
You can use a graphing calculator to find the z-score that
corresponds to a given area.
Use DISTR, invNorm or invcdf (area)
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 P98.
0.9803
–4 –3 –2 –1 0
1
2
3
4
z
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 (A16-17). Read the
values at the beginning of the corresponding row and at the top
of the column. The z-score is 2.06.
Finding z-Scores from Areas
Find the z-score corresponding to the 90th percentile (P90).
.90
0
DISTR, invNorm (.9) =
z
Finding z-Scores from Areas
Find the z-score corresponding to the 90th percentile (P90).
.90
0
z
DISTR, invNorm (.9) = 1.28. The z score is 1.28
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.
Finding z-Scores from Areas
Find the z-score with an area of .60 falling to its right.
.60
z
0
z
Finding z-Scores from Areas
Find the z-score with an area of .60 falling to its right.
.60
z
0
z
With .60 to the right, cumulative area is .40.
DIST, InvNorm(.4) =
Finding z-Scores from Areas
Find the z-score with an area of .60 falling to its right.
.60
z
0
z
With .60 to the right, cumulative area is .40.
DIST, InvNorm(.4) = -.25. 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
Finding z-Scores from Areas
Find the z-score such that 45% of the area under the
curve falls between –z and z.
.45
–z 0
z
Finding z-Scores from Areas
Find the z-score such that 45% of the area under the
curve falls between –z and z.
.45
–z 0
z
The area remaining in the tails is .55. Half this area is
in each tail so .55/2 = .275 is the cumulative area for
the -z value and .275 + .45 = .725 is the cumulative area
for the positive z. Use your calculator invNorm(.275) or
invNorm(.725) and the z-scores are 0.60 and -0.60.
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) 0
(b) –1.75
(c) 2.33
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 + (0)(7) = 152
(b) x = 152 + (–1.75)(7) = 139.75
(c) x = 152 + (2.33)(7) = 168.31
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?
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?
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.
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.
Section 5.4
The Central Limit
Theorem
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,
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
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
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.
69.2
Distribution of means of sample size 60,
will be normal.
mean
Standard deviation
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 > 30 the sampling distribution of will be normal
mean
standard deviation
Find the z-score for a sample mean of 70:
Interpreting the Central Limit Theorem
z
2.14
There is a 0.0162 probability that a sample of 60
men will have a mean height greater than 70”.
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 > 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.
Application Central Limit Theorem
P( 0.63 < z < 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.
Section 5.5
Normal Approximation
to the Binomial
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.
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.
Or…you 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
Why Do We Require np
0
1
2
3
4
5
5 and nq
n=5
p = 0.25, q = .75
np =1.25 nq = 3.75
n = 20
p = 0.25
np = 5 nq = 15
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
n = 50
p = 0.25
np = 12.5
nq = 37.5
0
10
20
30
40
50
5?
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
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.
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.
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.
.
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
approximation to the binomial.
5 use the normal
The binomial phrase of “fewer than 140” means
0, 1, 2, 3…139.
Use the correction for continuity to translate to the
continuous variable in the interval
. Find P( x <
139.5).
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 < 139.5).
P( z < -1.71) = 0.0436
The probability that fewer than 140 are in favor of
government regulation is approximately 0.0436.