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Yoonie is a personnel manager in a large corporation. Each month she must review 16of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2hours. Let 围 be the random variable representing the time it takes her to complete one review. Assume 围 is normally distributed. Let x-be the random variable representing the meantime to complete the 16reviews. Assume that the 16reviews represent a random set of reviews.

Find the probability that one review will take Yoonie from 3.5to 4.25hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability

b. P(________ <x< ________) = _______

Short Answer

Expert verified

From the information above

a) The probability that one review will take Yoonie from 3.5to 4.25hours is 0.2459. and the shaded region corresponding to the probability is:

b)p(3.5<x<4.25)=0.2459

Step by step solution

01

Given Information (part a)

It is provided that Yoonie reviews 16of the employees with an average time of 4hours and a standard deviation of 1.2hours. We have to find the probability that Yoonie will take 3.5to 4.25hours for one review and graph it.

02

Step 2: Explanation(part a)

According to the information, we have to find the probability that one review will take Yoonie from 3.5 to 4. 25

The Computation of the probability is given as below:

p(3.5<x<4.25)=p(Z<4.5)p(Z<3.5)

=p(Z<4.2541.2)p(Z<3.541.2)

=p(Z<0.21)p(Z<0.42)

Therefore,

localid="1648621489396" p(z<0.21)and

p(Z<0.42)in Excel is:

03

Step 3: Shaded region representation (part a)

Therefore,

p(3.5<x<4.25)=p(Z<0.21)p(Z<1.42)

=0.58320.3372

=0.2459

The shaded region corresponding to the probability is as shown below:

04

Given Information (part b)

It is provided that Yoonie reviews 16of the employees with an average time of 4hours and a standard deviation of1.2 hours

05

Explanation (part b)

The probability is

p(3.5<x<4.25)=p(Z<0.21)p(Z<1.42)

=0.58320.3372

Simplify,

localid="1648621737422" =0.2459

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