The wildlife department has been feeding a special food to rainbow trout fingerlings in a pond. Based on a large number of observations, the distribution of trout weights is normally distributed with a mean of 402.7 grams and a standard deviation 8.8 grams. What is the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams?
A) 0.3782
B) 0.0222
C) 1.0
D) 0.5

Question

Quizplus · Accepted Answer

The population mean is μ=402.7 grams and the standard deviation is σ=8.8 grams. The sample size is n=40. We need to find the probability that the sample mean weight, x̄, exceeds 405.5 grams. The distribution of x̄ is also normal with mean μ=μ=402.7 grams and standard deviation σ/√n=8.8/√40=1.392 grams. Thus, we need to standardize the sample mean:

Z=(x̄-μ)/(σ/√n)=(405.5-402.7)/(1.392)=1.997

Using a standard normal distribution table, we find the probability of Z being greater than 1.997 is 0.0222. Therefore, the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams is 0.0222 or 2.22%. Hence, the best choice is B.

The Wildlife Department Has Been Feeding a Special Food to Rainbow