The mean life of a television set is 121
months with a variance of 256
.
If a sample of 95
televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 2.2
months? Round your answer to four decimal places.
First, we calculate the standard deviation of the sample mean which is the square root of the variance divided by the sample size:
Standard deviation = √(256 / 95) = √2.6947368 ≈ 1.6417
Next, we calculate the standard error which is the standard deviation divided by the square root of the sample size:
Standard error = 1.6417 / √95 ≈ 0.1683
To find the probability that the sample mean would differ from the true mean by less than 2.2 months, we need to find the z-scores for both upper and lower limits:
Lower z-score = (121 - 2.2) / 0.1683 = 714.33
Upper z-score = (121 + 2.2) / 0.1683 = 722.88
Now, we look up the probabilities associated with these z-scores in a standard normal distribution table. Subtract the lower z-score probability from the upper z-score probability to get the final answer:
P(714.33 < Z < 722.88) ≈ P(Z < 722.88) - P(Z < 714.33) = 1 - P(Z < 714.33)
Using the standard normal distribution table or calculator, we find that P(Z < 714.33) ≈ 0.99999
Therefore, the probability that the sample mean would differ from the true mean by less than 2.2 months is approximately 0.00001.