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A variable is normally distributed with a mean 6and standard deviation 2. Find the percentage of all possible values of the variable that

a. lie between 1and 7

b. exceed 5

c. are less than 4.

Short Answer

Expert verified

(a) The percentage of all possible values of the variable that lie between 1and 7is 68.53%.

(b) The percentage of all possible values of the variable that exceed 5is 69.15%.

(c) The percentage of all possible values of the variable that are less than 4is15.87%.

Step by step solution

01

Part (a) Step 1: Given information

The given mean is 6

Standard deviation is2.

02

Part (a) Step 2: Explanation

The given data is written as

Mean, μ=6

Standard deviation,σ=2

The random variable Xhas a z-score which is given as

z=x-μσ

Find the z-scores

For 1,

z=1-62

=-2.5

For 7,

z=7-62

=0.5

Therefore the observations between 1and 7are the same as the z-scores between -2.5and 0.5.

The difference between the numbers in Table II is then used to calculate the percentage of all observations:

0.6915-0.0062=0.6853

03

Part (b) Step 1: Given information

The given mean is 6

Standard deviation is2.

04

Part (b) Step 2: Explanation

Use the above formula and find the z-score

z=5-62

=-0.5

As a result, observations more than 5correspond to z scores greater than -0.5.

The proportion of z-scores greater than -0.5is calculated in Table II of Appendix A

1-0.3085=0.6915

=69.15%.

05

Part (c) Step 1: Given information

The given mean is 6

Standard deviation is2.

06

Part (c) Step 2: Explanation

The z-score is obtained by

z=4-62

=-1.

Therefore the observations less than 4are the same z-scores less than -1. From Table II in Appendix A, the proportion of z-scores less than -1is

0.1587=15.87%.

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