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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 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. - .

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 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

- 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?

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.

- Draw a graph to represent this.

- What percentage of years does this investment

lose money, i.e. have an annual return of less

than 0? - 2.5

- What percentage of years is the annual return

more than 150? - 16

- 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 (s) 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 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.

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

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

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

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

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

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.

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(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.

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

Finding Normal Probabilities

- Example - continued

P(-.5 lt Z lt 1)

We need to find the shaded area

Finding Normal Probabilities

- Example - continued

P(-.5ltZlt0) P(0ltZlt1)

P(-.5ltZlt1)

P(0ltZlt1

.3413

z0

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)

Finding Normal Probabilities

- Example - continued

.3413

.1915

.5

-.5

Finding Normal Probabilities

- Example - continued

.3413

1.0

.5

-.5

P(-.5ltZlt1) P(-.5ltZlt0) P(0ltZlt1) .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 lt X lt 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 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.

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.

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.

Question 1

- If the magazines advertised mean sales per week

are correct, give a complete description of the

magazines weekly sales figures.

Answer 1

- 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 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?

Answer 2

- 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 magazines

advertised mean of 100,000 copies per week?

Explain

Answer 3

- 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.

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

contradictory to the conjecture.

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

- 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

- 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.