A survey is being planned to determine the mean amount of time corporation executives watch television. A pilot survey indicated that the mean time per week is 19 hours, with a standard deviation of 5.0 hours. It is desired to estimate the mean viewing time within 30 minutes. The 99% level of confidence is to be used. (Use z Distribution Table.)

Question

Copper

Mathematics

14 December, 22:54

A survey is being planned to determine the mean amount of time corporation executives watch television. A pilot survey indicated that the mean time per week is 19 hours, with a standard deviation of 5.0 hours. It is desired to estimate the mean viewing time within 30 minutes. The 99% level of confidence is to be used. (Use z Distribution Table.)

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Answers (1)

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Trevon Haley · Answer 1

We are given

x = 19

s = 5

u = 30

99% LOC

Getting the z-score

z = (19 - 30) / 5

z = - 2.2

From the z-score table,

the probability is

0.0139

In percentage, the answer is 0.39%