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## Normally Distributed Data

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### Chapter 4 Normally Distributed Data The Normal Distribution The normal distribution has these properties: It has a mean m, which defines its center. – PowerPoint PPT presentation

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Title: Normally Distributed Data

1
Chapter 4
• Normally Distributed Data

2
The Normal Distribution
• The normal distribution has these properties
• It has a mean m, which defines its center.
• It has a standard deviation s, which defines its
width.
• It is symmetric (mirror-image) about its mean m.
• It is bell-shaped.
• The total area under the curve is 1.
• The area to the left of m is 0.5 and equals the
area to the right of m.
• The STANDARD NORMAL distribution has m 0 and s
1.
• .

3
The effects of m and s
How does the standard deviation affect the shape
of f(x)?
s 2
s 3
s 4
How does the expected value affect the location
of f(x)?
m 10
m 11
m 12
4
The Normal Curve Rule
• For any normal distribution with a mean m and a
standard deviation s, the following statements
are true
• Approximately 68 of all observations fall within
one standard deviation of the mean, or in the
interval (m s, m s).
• Approximately 95 of all observations fall within
two standard deviations of the mean , or in the
interval (m 2s, m 2s).
• Almost all observations fall within three
standard deviations of the mean, or in the
interval (m 3s, m 3s).

5
Example
• Pulse rates (the number of times a persons heart
beats per minute) are normally distributed with a
mean of 80 and a standard deviation of 6.
• Draw a graph to represent this.

80
92
86
98
74
68
62
6
• What percentage of pulse rates is between 68 and
92 beats per minute?
• 95 Why?
• What percentage of pulse rates is more than 86
beats per minute?
• 16 Why?
• A woman reports that only 2 ½ of all peoples
pulse rates are lower than her pulse rate. What
is the womans pulse rate?
• 68 Why?

7
Another Example
• From past experience it is known that for a
particular investment the amount of money earned
annually on a 1000 investment averages 100.
The amount earned is called the annual return and
it varies from year to year. The annual return
is known to be normally distributed with a
standard deviation of 50.

8
• Draw a graph to represent this.

9
• What percentage of years does this investment
lose money, i.e. have an annual return of less
than 0?
• 2.5

10
• What percentage of years is the annual return
more than 150?
• 16

11
• A person who invests 1000 would like to have a
total amount of money of at least 1150 by next
year. Do you think this will happen? Explain
fully.
• This is not unlikely. It happens 16 of the
time. However, this is not a large enough
percentage to assure that the person will have
1150.

12
Normal Distribution
interest, X, which has a normal distribution with
parameters m and s.
• We then transform the question asked into one
involving the standard normal distribution. This
is the power of the normal distribution, which
only requires one table to do all exercises.
• We begin by transforming from the given
population into the standard normal curve values
by creating z-scores.
• A z-score represents how many standard deviations
above or below the mean a raw score is located.

13
Normal Distribution
• But what if our problem does not give us the z
score? How do we find it?
• There is an equation that we can use
• X the measurement we have from the problem.

14
Z-scores
• A z-score is calculated as z
• A positive z-score means the value of x is above
the mean.
• A negative z-score means the value of x is below
the mean.
• A z-score of 0 means the value of x is equal to
m, the mean.

15
Normal Distribution
• This equation says to take the measurement we
have (x) , subtract the mean (µ) from it to get
the distance from µ,
• Then divide this distance by the standard
deviation (s) to find the number of standard
deviations that this distance represents.
• Z the number of standard deviations away from
the mean (µ)

16
Normal Distribution
• Say we have a measurement of 10 seconds. We know
the mean is 7 seconds, and the standard deviation
is 2 seconds. The z score then is
• 10 7 3
• 3 / 2 1.5
• The measurement is 1.5 standard deviations away
from the mean of 7.

17
Z-scores
• Suppose the monthly utility bill for residents in
the area during June follows a normal
distribution with a mean of 100 and a standard
deviation of 15.
• Find and interpret the z-score for a bill of
• a) 80 b) 133
• z -1.33
z 2.20
• A bill of 80 is 1.33 standard A bill of 133 is
2.20
• deviations below the mean standard deviations
above
• utility bill amount. the mean utility bill
amount.

18
The Standard Normal Distribution
• The table in the back of your book gives P(0 lt Z
lt a) for any positive number a.
• What is P(0 lt Z lt 2.15)? We are interested in
the area under the standard normal curve between
0 and 2.15 on the horizontal number line.
• Be aware of the difference between the z-score
values on the horizontal axis and the area under
the curve.
• Keys to answering probability questions
• DRAW PICTURES!
• Use pictures including 0 in the range to make
puzzle pieces which fit together to give you the
picture you want.

19
Using the Standard Normal Table
Standard normal probabilities have been
calculated and are provided in a table .
The tabulated probabilities correspond to the
area between Z0 and some Z z0 gt0
Z z0
Z 0
20
Standard Normal Distribution Example 1
• What is P(0 lt Z lt 2.15)? So, we look for z 2.1
0.05 in the back of our book. Hence, P(0 lt Z lt
2.15) 0.484

21
Standard Normal Distribution Example 2
• What is P(0 lt Z lt 1.26)? So, we look for z 1.2
0.06 in the back of our book. Hence, P(0 lt Z lt
1.26) 0.396

22
Normal Distribution Steps
• In general, the steps in finding a probability
associated with data which comes from a normal
distribution are
• Identify the mean and standard deviation for the
data
• Identify the values of interest
• convert the values of interest into z-scores
• use the normal curve areas table
• summarize the results

23
Normal Distribution Example 1
• The rate of return (X) on an investment is
normally distributed with mean of 10 and
standard deviation of 5
• What is the probability of losing money?

X
0 - 10 5
.4772
(i) P(Xlt 0 ) P(Zlt ) P(Zlt - 2)
Z
2
0
-2
P(Zgt2)
0.5 - P(0ltZlt2) 0.5 - .4772 .0228
24
Normal Distribution Example 2
• Suppose the amount of soda in 12 oz cans of Diet
Coke is normally distributed with m 12.05 oz
and s 0.05 oz.
• a) What is the probability a randomly selected
can has less than 12 ounces of soda?
• Solution Let X amount of soda in a randomly
selected can.
• P(X lt 12) is equivalent to the following
• But P(Z lt -1.00) 0.5 0.341 0.159.
• So the answer to P(X lt 12) is 0.159.

25
Example 3 The amount of time it takes to assemble
a computer is normally distributed, with a mean
of 50 minutes and a standard deviation of 10
minutes. What is the probability that a computer
is assembled in a time between 45 and 60 minutes?
26
Finding Normal Probabilities
• Solution
• If X denotes the assembly time of a computer, we
seek the probability P(45ltXlt60).
• This probability can be calculated by creating a
new normal variable the standard normal variable.

Every normal variable with some m and s, can be
transformed into this Z.
Therefore, once probabilities for Z are
calculated, probabilities of any normal variable
can be found.
27
Finding Normal Probabilities
• Example - continued

- m
45
X
60
- 50
- 50
P(45ltXlt60) P( lt lt
)
s
10
10
P(-0.5 lt Z lt 1)
To complete the calculation we need to compute
the probability under the standard normal
distribution
28
Finding Normal Probabilities
• Example - continued

P(-.5 lt Z lt 1)
We need to find the shaded area
29
Finding Normal Probabilities
• Example - continued

P(-.5ltZlt0) P(0ltZlt1)
P(-.5ltZlt1)
P(0ltZlt1
.3413
z0
30
Finding Normal Probabilities
• The symmetry of the normal distribution makes it
possible to calculate probabilities for negative
values of Z using the table as follows

-z0
z0
0
P(-z0ltZlt0) P(0ltZltz0)
31
Finding Normal Probabilities
• Example - continued

.3413
.1915
.5
-.5
32
Finding Normal Probabilities
• Example - continued

.3413
1.0
.5
-.5
P(-.5ltZlt1) P(-.5ltZlt0) P(0ltZlt1) .1915 .3413
.5328
33
Normal Distribution Example 4
• Suppose the amount of soda in 12 oz cans of Diet
Coke is normally distributed with m 12.05 oz
and s 0.05 oz.
• b) What is the probability a randomly selected
can contains between 11.98 and 12.03 ounces of
soda?
• So P(11.98 lt X lt 12.03) is equivalent to

34
Normal Distribution Example 5
• Suppose the amount of hours worked per week for
full time students follows a normal distribution
with m 20 hours and s 4.5 hours.
• What is the probability that a randomly selected
full time student works
• a) more than 26 hours during one week?
• b) between 15 and 22 hours during one week?
• c) less than 15 hours during one week?
• d) between 25 and 30 hours during one week?

35
Normal Distribution Backwards Problems
• These exercises give you a probability (often as
a ), and ask you to find the value of a which
makes a probability statement true. (ie P(X gt a)
0.90 for what value of a?)
• Example
• Suppose the amount of hours worked per week for
full time students follows a normal distribution
with m 20 hours and s 4.5 hours.
• 90 of all students work less than how many hours
per week?
• Here, P(X lt a) 0.90 (from 90).
• We want to find a, but we must begin with the
standard normal distribution.

36
Normal Distribution Backwards Problems
• P(Z lt b) 0.90 when P(0 lt Z lt b) 0.90 0.5
0.40. Look for 0.40 in the body of your normal
probability table.
• You should find b 1.28 (yes, it is positive)
• Now recall Z but b Z and X a for
our exercise.
• Then 1.28 , so a 1.28(4.5) 20
25.76 hours.
• So, 90 of all full time students work less than
25.76 hours per week.

37
Using the Normal Distribution to make inferences.
• A weekly magazine advertises that the mean number
of copies sold per week is 100,000. It is known
that the distribution of weekly sales is normally
distributed with a standard deviation of 3,500
copies. We suspect that the magazine actually
sells fewer copies than advertised. To see if
the magazines conjectured mean is wrong, we
randomly select one week and observe the
magazines sales.

38
Question 1
• If the magazines advertised mean sales per week
are correct, give a complete description of the
magazines weekly sales figures.

39
• The distribution should be a normal shape, with
the mean of 100,000, and z marks at 89.5, 93.0,
96.5, 100, 103.5, 107, and 110.5 (thousands)

40
Question 2
• Suppose that when one weeks sales of the
magazine are randomly sampled, there were 90,000
copies sold. If the advertised mean is correct,
what is the probability that one weeks sales
would be 90,000 copies or less?

41
• The z value for 90,000 would be -2.86
• (90,000 100,000)/3,500
• The p-value for -2.86 is
• 0.5000 0.4979 0.0021

42
Question 3
• Would observing the week with sales of 90,000
copies be contradictory to the magazines
advertised mean of 100,000 copies per week?
Explain

43
• If the advertised mean were correct, there is a
0.21 probability that we could observe a week
with sales of 90,000 or less. This is not much
probability.
• This observation seems to contradict the original
claim of 100,000 copies per week.

44
Probability rule
• If the probability of an observed sample is 0.05
or less, assuming the truth of some conjecture,
then the sample is contradictory to that
conjecture. Otherwise, the sample is not

45
Inference Making Procedure
• Let µ be the conjectured value of the mean of a
normal population with standard deviation s. Let
x be a randomly sampled value from this
population. To see if the sampled value of x is
contradictory to the conjectured mean, complete
the following steps

46
• 1. Sketch the normal distribution assuming that
the conjectured value of µ is correct. Mark the
observed value of x on your sketch.
• 2. Find the z-score corresponding to the
observed value of x using

47
• 3. Using the p-value table, find the probability
of observing a value from the distribution that
is as far from the mean as x, or further.
• 4. If the probability found in Step 3 is less
than 0.05, the observed value of x is
contradictory to the conjectured value of the
mean. Otherwise, it is not contradictory.