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.
Find the probability that the mean of a month's reviews will take Yoonie from 3.5 to 4.25 hrs.
a. Give the probability statement and the probability. (Enter exact numbers as integers, fractions, or decimals for the probability statement. Round the probability to four decimal places.
Answers
Answer 1
The probability that the mean of a month's reviews will take Yoonie from 3.5 to 4.25 hrs is 0.7499.
What is the probability?We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 3.5
x2 = upper bound = 4.25
u = mean = 4
n = sample size = 16
s = standard deviation = 1.2
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -1.666666667
z2 = upper z score = (x2 - u) * sqrt(n) / s = 0.833333333
Using table/technology, the left-tailed areas between these z scores is
P(z < z1) = 0.047790352
P(z < z2) = 0.797671619
Thus, the area between them, by subtracting these areas, is
P(3.5<xbar<4.2) = 0.749881267
The probability is 0.7499.
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