Question

The mean life of a television set is 96 months with a variance of 225.

If a sample of 76 televisions is randomly selected, what is the probability that the sample mean would be less than 94.7 months? Round your answer to four decimal places.

Answer 1

To find the probability that the sample mean is less than 94.7 months, calculate the standard error of the mean, find the Z-score, and then look up the probability corresponding to that Z-score in a standard normal distribution.

Step-by-step explanation:

The student is asking about the probability that the sample mean of television lifespans is less than 94.7 months, given a population mean of 96 months and a population variance of 225. Since the sample size is 76 televisions, you can use the Central Limit Theorem to approach this problem by treating the distribution of the sample mean as a normal distribution (since the sample size is large).

First, find the standard error of the mean (SEM) which is the standard deviation of the sampling distribution of the sample mean:

SEM = √(variance / sample size) = √(225 / 76) ≈ 1.72 months

Next, calculate the Z-score for the sample mean of 94.7 months:

Z = (sample mean - population mean) / SEM = (94.7 - 96) / 1.72 ≈ -0.7558

Finally, find the probability that the Z-score is less than -0.7558 using a standard normal distribution table or a calculator. This gives you the probability that the sample mean is less than 94.7 months.