qbm117 business statistics probability and probability distributions the normal distribution...

QBM117Business Statistics

Probability and Probability Distributions

The Normal Distribution continued

1

Objectives

• To learn how to use the Z tables in reverse to find the value of Z corresponding to a known probability.

• To learn how to use the Z tables in reverse to find the value of X corresponding to a known probability.

2

Finding Values That Correspond to Known Probabilities

• Start by looking on the inside of the table for the known probability.

• Then move to the outside of the table to determine the associated z value.

• Then back transform to obtain the associated x value.

3

X Z

Example 1An area of 0.4370 lies under the standard normal curve between the mean and a given positive z score. What is the value of that z score?

We want to find such that

The z value corresponding to the area of 0.4370 is

4

*z *(0 ) 0.4370P Z z

0 *z

0.4370

Z

* 1.53z

An area of 0.25 lies under the standard normal curve between the mean and a given positive z score. What is the value of that z score?

We want to find such that

Example 2

5

*z *(0 ) 0.25P Z z

0 *z

0.25

Z

To find we search the table for the probability 0.25.

We don’t find this probability but we find two that are close: 0.2486 and 0.2517.

0.25 is closer to 0.2486 than it is to 0.2517.

And so we look up the z value associated with 0.2486, which is * 0.67z

*z

Example 3An area of 0.05 lies under the standard normal curve above a given positive z score. What is the value of that z score?

We want to find such that

7

*z *( ) 0.05P Z z

Z*z0

0.05

Hence we want to find such that

8

*z

*(0 ) 0.5 0.05

0.45

P Z z

Z*z0

0.05

0.45

To find we search the table for the probability 0.45.

We don’t find this probability but we find two that are close: 0.4495 and 0.4505

0.45 is exactly half way between 0.4495 and 0.4505.

And so we look up the z value associated with both probabilities and average them.

The Z values associated with these probabilities are 1.64 and 1.65.

The average of these values is 1.645, hence * 1.645z

*z

9

Exercise 1

Find the value for which

a.

b.

c.

10

*z

*( ) 0.15P Z z

*( ) 0.80P Z z

* *( ) 0.90P z Z z

Example 3

Scores of an aptitude test given by a training department of a large company are normally distributed with a mean of 75 points and a standard deviation of 5 points.The company has decided that people who score in the bottom 10% of the test scores will not receive any additional job training. If there are to be layoffs, these people will be among the first to be cut. What cut-off score on the test should the company use?

11

Let X = the score on the aptitude test

We want to find the value of X that has 10% below it.

12

2~ (75,5 )X N

0.10

X75*x

First we find the value of Z that has 10% below it and then we back transform to find the corresponding value of X.

We want to find such that

Using the Z tables we find 13

0.10

Z0*z

0.40

*z *( ) 0.10P Z z

* 1.28z

We now need to back transform to find

Therefore the company should set the cut-off at 68.6 points.

14

*x

* *

75 ( 1.28) 5

68.6

x z

Example 3 continued

The company is planning to give extra training to employees who score in the top 2% of those taking the test. The company would like to identify the score to use as the cut-off point.

15

Recall that X = the score on the aptitude test

We want to find the value of X that has 2% above it.

16

2~ (75,5 )X N

X75 *x

0.02

First we need to find the value of Z that has 2% above it.

From the Z tables we find

Hence

Therefore the cut-off should be 82.25 points.

Z0 *z

0.02

0.48

* 2.05z

* 75 2.05 5

85.25

x

Exercise 2

The marks for a first year statistics exam are normally distributed with a mean of 72 and a standard deviation of 14.

Suppose the lecturer wants to assign High Distinctions to the top 15% and fail the bottom 20%.

a. What is the cut-off score for a HD?

b. What is the pass mark?

Exercise 3

The number of pages printed before replacing the cartridge in a laser printer is normally distributed with a mean of 11500 pages and a standard deviation of 800 pages. The manufacturer wants to provide guidelines to potential customers advising them the minimum number of pages they can expect from each cartridge. How many pages should it advertise if the company wants to be correct 99% of the time?

Exercise 4

The Rural Bank is reviewing its service charges and interest-paying policies on cheque accounts. The bank has found that the average daily balance on personal cheque accounts is normally distributed with a mean of $550.00 and a standard deviation of $150.00.

a. What percentage of personal cheque account customers carry average daily balances below $200?

b. The bank is considering paying interest to customers carrying average daily balances in excess of a certain amount. If the bank does not want to pay interest to more than 8% of its customers, what is the minimum average daily balance it should be willing to pay interest on?

Exercises

• 5.33• 5.79

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