The Normal Distribution

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The Normal Distribution
Chapter 5
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Properties of a Normal Distribution

• The mean, median, and mode are equal

• Bell shaped and symmetric about the mean

• The total area under the curve is one (1) or 100

• The curve approaches but never touches the x-
  axis as it
• extends farther and farther away from the mean in
  both directions.

• The points at which the curvature changes are
  called inflection points.

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Means and Standard Deviations
Curves with different means, same standard
deviation
Curves with different means different standard
deviations
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Empirical Rule
5
Determining Intervals
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
hours. 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.
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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
standard deviation of 7. Find the standard
z-score for a person with a score of (a) 161
(b) 148 (c) 152
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From z-Scores to Raw Scores
The test scores for a civil service exam are
normally distributed with a mean of 152 and
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
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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
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Cumulative Areas
The total area under the curve is one.

• 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.

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Cumulative Areas
Find the cumulative area for a z-score of -1.25.
Read down the z column on the left to z -1.2
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. P ( z ? -1.25) 0.1056
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From Areas to z-scores
Find the z-score corresponding to a cumulative
area of 0.9803.
0.9803
0.9803
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.
z 2.06 is roughly the 98th percentile.
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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 -1.24)
Read down the z-column to -1.2 and across to .04.
The cumulative area is 0.1075.
P ( z lt 1.24) 0.1075
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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
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
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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)
1. P(z lt 1.17) 0.8790
2. P(z lt -1.25) 0.1056
3. P( -1.25 lt z lt 1.17) 0.8790 - 0.1056 0.7734
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Summary
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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 standard deviation of 15. Find the
probability that a person selected at random will
have an IQ score less than 115.
To find the area in this interval, first find the
standard score equivalent to x 115.
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Probabilities and Normal Distributions
Standard Normal Distribution
Find P(z lt 1)
P( z lt 1) 0.8413, so P( x lt115) 0.8413
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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.
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Finding Percentiles
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?
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.
x 100 1.28(12) 115.36.
115.36 is the smallest value for the top 10.
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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
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The Central Limit Theorem
the sample means will have a normal distribution
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The Central Limit Theorem
x
the distribution of means of sample size n , will
be normal with a mean standard deviation
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Application
Distribution of means of sample size 60 , will
be normal.
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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.
mean
Find the z-score for a sample mean of 70
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Interpreting the Central Limit Theorem
P (z gt 2.14) 1 - 0.9838 0.0162
z
There is a 0.0162 probability that a sample of 60
men will have a mean height greater than 70.
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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.
standard deviation
Calculate the standard z-score for sample values
of 1.169 and 1.179.
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Application Central Limit Theorem
P( 0.63 lt z lt 1.90) 0.9713 - 0.7357 0.2356
The probability is 0.2356 that the mean for the
sample is between 1.169 and 1.179.
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Normal Approximations to the Binomial
Characteristics of a Binomial Experiment

• There are a fixed number of trials. (n)
• The n trials are independent and repeated under
  identical conditions
• 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
• The central problem is to find the probability of
  x successes out of n trials. Where x 0 or 1 or
  2 n.

x is a count of the number of successes in n
trials.
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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
301exactly 500 Americans have A blood type
and add the probabilities.
Oryou could use the normal curve probabilities
to approximate the binomial probabilities.
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Why do we require np ? 5 and nq ? 5?
n 5 p 0.25, q .75 np 1.25 nq 3.75
n 20 p 0.25 np 5 nq 15
4
n 50 p 0.25 np 12.5 nq 37.5
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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 P (14? x ? 16)
Add the areas of the rectangles with midpoints at
x 14, x 15, x 16.
0.111 0.089 0.065 0.265
P (14? x ? 16) 0.265
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Correction for Continuity
Use the normal approximation to the binomial to
find P(14? x ? 16) if n 50 and p 0.25
Check that np 12.5 ? 5 and nq 37.5 ? 5.
The interval of values under the normal curve is
13.5 ? x ? 16.5.
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.
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Normal Approximation to the Binomial
Use the normal approximation to the binomial to
find P(14? x ? 16) if n 50 and p 0.25
Find the mean and standard deviation using
binomial distribution formulas.
Adjust the endpoints to correct for continuity
P(13.5 ? x ? 16.5)
Convert each endpoint to a standard score
P(0.33 ? z ? 1.31) 0.9049 - 0.6293 0.2756
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Application
A survey of Internet users found that 75 favored
government regulations on junk e-mail. If 200
Internet users are randomly selected, find the
probability that fewer than 140 are in favor of
government regulation.
Since np150 ? 5 and nq 50 ? 5 use the normal
approximation to the binomial.
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 0.0436