Download Chapter 5: The Normal Distribution

Chapter 5: The Normal Distribution
MODELING CONTINUOUS VARIABLES
Histogram 6.1
Proportion
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0.08
0.06
0.04
0.02
30
35
40
45
50
55
Age
If we draw a curve through the tops of the bars in Histogram 6.1 and require the smoothed
curve to have total area under it equal to 1, we would have what is called a density
function, also called a density curve. The key idea when working with density functions
is that area under the curve, above an interval, corresponds to the proportion of units with
values in the interval.
NOTATION...
Since we will be discussing models for populations, the mean and standard deviation for a
density curve or model will be represented by
 (mu) and 
(sigma), respectively.
DEFINITION:
A density function is a (nonnegative) function or curve that describes the overall shape of
a distribution. The total area under the entire curve is equal to 1, and proportions are
measured as areas under the density function.
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Normal Distributions
General Notation
X is N(  , ) means
that
the
variable
or
characteristic X is normally distributed with mean


Threeand
members
of the family
of normal .distributions
standard
deviation
Distribution #3:
Normal with a mean of 80
and a standard deviation of 5
Distribution #1:
Normal with a mean of 50
and a standard deviation of 10
20
30
40
50
Distribution #2:
Normal with a mean of 80
and a standard deviation of 10
60
70
80
90
100
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APPLY THE EMPIRICAL RULE
Example 1 IQ Scores
Let the variable X represent IQ scores of 12-year-olds. Suppose that the distribution of X is
normal with a mean of 100 and a standard deviation of 16—that is, X is
N(100, 16). Jessica is a 12-year-old and has an IQ score of 132.We would like to determine
the proportion of 12-year olds that have IQ scores less than Jessica’s score of 132.
Since the area under the density curve corresponds to proportion, we want to find the area
to the left of 132 under an N(100, 16) curve. Sketch this curve and show the corresponding
area that represents this proportion.
IQ Scores have a
normal distribution
with mean 100 and
standard deviation 16

area to the left of 132 = ?
68
84
100
116
132
IQ Score
Let's Do It!
A physical fitness association is including the mile run in its secondary-school fitness test.
The time for this event for boys in secondary school is known to possess a normal
distribution with a mean of 460 seconds and a standard deviation of 50 seconds. Between
what times do we expect approximately 95% of the boys to run the mile?
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STANDARD NORMAL DISTRIBUTION AND Z-SCORES
DEFINITION:
If
X is N(  , ) ,
the standardized normal variable
Z
X

is
N 0 ,1.
DEFINITION:
The z-score or standard score for an observed value tells us how many standard
deviations the observed value is from the mean – that is, it tells us how far the observed
value is from the mean in standard-deviation units. It is computed as follows:
Z
X

= number of standard deviations that
X differs from the mean

If Z > 0, then the value of X is above (greater than) its mean.
If Z < 0, then the value of X is below (less than) its mean.
If Z = 0, then the value of X is equal to its mean.
Example 2 Standard IQ Score
Recall the distribution of IQ scores for 12-year-olds—normally distributed with a mean of
100 and a standard deviation of 16.
(a)
Jessica had a score of 132. Compute Jessica’s standardized score.
(b)
Suppose Jessica has an older brother, Mike, who is 20 years old and has an IQ
score of 144. It wouldn’t make sense to directly compare Mike’s score of 144 to
Jessica’s score of 132. The two scores come from different distributions due to the
age difference. Assume that the distribution of IQ scores for 20-year-olds is normal
with a mean of 120 and a standard deviation of 20. Compute Mike’s standardized
score.
(c)
Relative to their respective age group, who had the higher IQ score—Jessica or
Mike?
Solution
Jessica's standard score =
Mike's standard score =
132  100
 2.
16
144  120
 12
. .
20
Thus, relative to their respective age groups, Jessica has a higher IQ score than Mike

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Method of Z-scores for Identifying Outliers
Rules for detecting outliers:
Box-Plot: outliers are data points beyond the lower and upper fences
Z-score: outliers have a z-score that is more than 3 standard deviations from the
mean, |z| > 3.
Let's Do It! 6
At the U.S. Open Tennis Championship a statistician keeps track of every serve that a
player hits during the tournament. The statistician reported that the mean serve speed of a
particular player was 103 miles per hour (mph) and the standard deviation of the serve
speeds was 12 mph. Using the z-score approach for detecting outliers, which of the
following serve speeds would represent outliers in the distribution of the player's serve
speeds?
61 mph:
115 mph:
and 127 mph:
Let's Do It!
Suppose female bank employee believes that her salary is low as a result of sex
discrimination. To substantiate her belief, she collects information on salaries of her male
counterparts in the banking business. She finds that their salary is approximately normal
with an average of $54,000 and a standard deviation of $2000. Her salary is $47,000. Does
this information support her claim of sex discrimination?
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How to Calculate Areas under a Normal Distribution
Finding Proportions for the Standard Normal Distribution
Finding proportions under a normal distribution involves standardization and then finding
the corresponding proportion (area) under the standard normal distribution. Let’s first work
on finding areas under a standard normal N(0, 1) distribution.
(a)
Find the area under the standard normal distribution to the left of z = 1.22. Sketch
a picture of the corresponding area and use your TI to find the area.
Solution
The area to the left of is described below. z = 1.22

area to the
left of z=1.22
is 0.8888
0
z=1.22
Z
Using TI:
(b)
Find the area under the standard normal distribution to the right of z = 1.22.
Solution
We already know the area to the left of z = 1.22, and we know that the total area
under any density to he right of 0 is equal to 0.5. So the area to the right of z =
1.22 must be 0.5 - 0.3888 = 0.1112. We could also find the area to the left of z =
-1.22. By the symmetry of the standard normal distribution, the area to the right
of z = 1.22 is equal to the area to the left of z = -1.22.
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Let's Do It!More Standard Normal Areas
a. Find the area under the normal distribution curve to
the right of z = 1.38. Round to four decimal places.
b. Find the area under the normal distribution curve between z 1 = 2.00 and z 2 = 2.55.
Round your answer to four decimal places.
c. Use the standard normal distribution to find P(-2.25 < z < 0).
d. For a standard normal random variable, find the probability that z exceeds the value
-1.65.
e. Obtain the shaded area under the standard normal curve.
Calculate Areas under a General Normal Distribution( Non-Standard)
Find the indicated probability or percentage for the normally distributed variable.
The variable X is normally distributed. The mean is μ = 60.0 and the standard deviation is σ
= 4.0. Find P(X < 53.0).
Let's Do It!
Physical fitness association is including the mile run in its secondary-school fitness test.
The time for this event for boys in secondary school is known to possess a normal
distribution with a mean of 440 seconds and a standard deviation of 50 seconds. Find the
probability that a randomly selected boy in secondary school can run the mile in less than
325 seconds.
Let's Do It!
The weight of corn chips dispensed into a 24-ounce bag by the dispensing machine has
been identified as possessing a normal distribution with a mean of 24.5 ounces and a
standard deviation of 0.2 ounce. What proportion of the 24-ounce bags contain more than
the advertised 24 ounces of chips?
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Let's Do It!
A bank's loan officer rates applicants for credit. The ratings are normally distributed with a
mean of 200 and a standard deviation of 50. If an applicant is randomly selected, find the
probability of a rating that is between 200 and 275.
Finding Percentile scores, top and bottom scores of a the Normal Distribution
Example 
The Top 1% of the IQ Distribution


Recall the N 100,16 model for IQ score of 12-year-olds. What IQ score must a 12-yearold have to place in the top 1% of the distribution of IQ scores? Find the percentile using
your calculator.
Again it may be helpful to draw a picture:

The area to the left is 0.99
100
?
IQ Score
Many calculators have the ability to find various percentiles of a normal distribution. The TI has a
built-in function called invNorm under the DIST menu. You must first specify the desired are to the
left, then the mean and the standard deviation for the normal distribution. The steps for finding the
99th percentile of our N(100,16) distribution are as follows:
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Let's Do It!
a. The lifetimes of lightbulbs of a particular type are normally distributed with a mean of
235 hours and a standard deviation of 7 hours. Find the first quartile.
b. The weights of certain machine components are normally distributed with a mean of
8.83 g and a standard deviation of 0.1 g. find the 97th percentile.
c. The board of examiners that administers the real estate broker's examination in a
certain state found that the mean score on the test was 435 and the standard
deviation was 72. If the board wants to set the passing score so that only the best
10% of all applicants pass, what is the passing score? Assume that the scores are
normally distributed.
Let's Do It!
a. Find a value of the standard normal random variable z, called z0, such that P(z ≤z0)
= 0.70.
b. Find a value of the standard normal random variable z, called z0, such that
P(-z0≤ z ≤ z0) = 0.9234.
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