Question

Yoonie is a personnel manager in a large corporation. Each month she must review 16 of 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 1.2 hours. Let X be the random variable representing the time it takes her to complete one review. Assume X is normally distributed. Let X be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews. Complete the distributions. (Enter exact numbers as integers, fractions, or decimals.)

Answer 1

Answer:

(a) The probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hours is 0.7492.

(b) The values representing the 95th percentile for the mean time to complete one month's reviews is 4.50 hours.

Step-by-step explanation:

The random variable X is defined as the time it takes her to complete one review.

The random variable X follows a Normal distribution with parameters μ = 4 hours and σ = 1.2 hours.

A random sample of n = 16 reviews are selected as a set.

(a)

Compute the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hours as follows:

$P(3.5$

                              $=P(-1.67$

Thus, the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hours is 0.7492.

(b)

The pth percentile is a data value such that at least p% of the data set is less than or equal to this data value and at least (100 - p)% of the data set are more than or equal to this data value.

So, the 95th percentile for the mean time to complete one month's reviews can be represented as:

$P(\bar X$

This implies that:

$P(\bar X$

The value of z is:

z = 1.645

Compute the value of a as follows:

$z=\frac{a-4}{1.2/\sqrt{16}}\\1.645=\frac{a-4}{0.3}\\z=4+(0.3\times 1.645)\\z=4.4935\\z\approx4.50$

Thus, the values representing the 95th percentile for the mean time to complete one month's reviews is 4.50 hours.