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 of 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

We can use the central limit theorem to approximate the sampling distribution of the sample mean. Since the sample size is 40, which is considered relatively large, we can safely assume that the sampling distribution is approximately normal with a mean of 402.7 grams and a standard deviation of 8.8/√40 = 1.392 grams.

To find the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams, we need to standardize the sample mean using the formula:

z = (x̄ - μ) / (σ / √n)

where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

Plugging in the values, we get:

z = (405.5 - 402.7) / (1.392) = 1.998

Using a standard normal table, we can find the probability that a z-score is greater than 1.998. The table gives us a probability of 0.0222, so the answer is B)0.0222.

The Wildlife Department Has Been Feeding a Special Food to Rainbow