Question

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.)

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%