Chapter 4

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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. It has a standard deviation s, which defines its width. It is symmetric (mirror-image) about it’s mean m . It is bell-shaped.

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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.

It has a standard deviation s, which defines its width.

It is symmetric (mirror-image) about it’s 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.

.

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

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).

Example
• Pulse rates (the number of times a person’s 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

62

68

74

86

98

• 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 woman’s pulse rate?
• 68 Why?
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.
What percentage of years does this investment lose money, i.e. have an annual return of less than \$0?
• 2.5%
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.

Normal Distribution

Typically, we start with a random variable of 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.

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.

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.

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 (σ) to find the number of standard deviations that this distance represents.
• Z = the number of standard deviations away from the mean (μ)
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.

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.

The Standard Normal Distribution

The table in the back of your book gives P(0 < Z < a) for any positive number a.

What is P(0 < Z < 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.

DRAW PICTURES!

Use pictures including 0 in the range to make puzzle pieces which fit together to give you the picture you want.

P(0<Z<z0)

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 Z=0 and some Z = z0 >0

Z = z0

Z = 0

Standard Normal Distribution – Example 1

What is P(0 < Z < 2.15)? So, we look for z = 2.1 + 0.05 in the back of our book. Hence, P(0 < Z < 2.15) = 0.484

Standard Normal Distribution – Example 2

What is P(0 < Z < 1.26)? So, we look for z = 1.2 + 0.06 in the back of our book. Hence, P(0 < Z < 1.26) = 0.396

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

0%

10%

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(X< 0 ) = P(Z< ) = P(Z< - 2)

Z

2

-2

0

=P(Z>2) =

0.5 - P(0<Z<2) = 0.5 - .4772 = .0228

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 < 12) is equivalent to the following:

=

But P(Z < -1.00) = 0.5 – 0.341 = 0.159.

So the answer to P(X < 12) is 0.159.

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?

Finding Normal Probabilities

• Solution
• If X denotes the assembly time of a computer, we seek the probability P(45<X<60).
• 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.

Finding Normal Probabilities

• Example - continued

- m

45

- 50

X

60

- 50

P(45<X<60) = P( < < )

s

10

10

= P(-0.5 < Z < 1)

To complete the calculation we need to compute

the probability under the standard normal distribution

- m

45

- 50

X

60

- 50

P(45<X<60) = P( < < )

s

10

10

z0 = 1

z0 = -.5

Finding Normal Probabilities

• Example - continued

= P(-.5 < Z < 1)

We need to find the shaded area

- m

45

- 50

X

60

- 50

P(45<X<60) = P( < < )

s

10

10

z0 = 1

z0 =-.5

Finding Normal Probabilities

• Example - continued

= P(-.5<Z<1) =

P(-.5<Z<0)+ P(0<Z<1)

P(0<Z<1

.3413

z=0

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(-z0<Z<0) = P(0<Z<z0)

Finding Normal Probabilities

• Example - continued

.3413

.1915

-.5

.5

.1915

.1915

.1915

.1915

Finding Normal Probabilities

• Example - continued

.3413

1.0

-.5

.5

P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = .1915 + .3413 = .5328

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 < X < 12.03) is equivalent to

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?

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 > 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 < a) = 0.90 (from 90%).

We want to find “a”, but we must begin with the standard normal distribution.

Normal Distribution – Backwards Problems

P(Z < b) = 0.90 when P(0 < Z < 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.

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 magazine’s conjectured mean is wrong, we randomly select one week and observe the magazine’s sales.
Question 1
• If the magazine’s advertised mean sales per week are correct, give a complete description of the magazine’s weekly sales figures.
• 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)
Question 2
• Suppose that when one week’s 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 week’s sales would be 90,000 copies or less?
• 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
Question 3
• Would observing the week with sales of 90,000 copies be contradictory to the magazine’s advertised mean of 100,000 copies per week? Explain