Question

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? Select one: a. 0.3782 b. 0.0222 c. 1.0 d. 0.5

Answer 1

b. 0.0222

Explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`(\sigma)/(√(n))`.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 402.7, \sigma = 8.8, n = 40, s = (8.8)/(√(40)) = 1.39`

What is the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams?

This is 1 subtracted by the pvalue of Z when X = 405.5. So

`Z = (X - \mu)/(s)`

`Z = (405.5 - 402.7)/(1.39)`

`Z = 2.01`

`Z = 2.01` has a pvalue of 0.9778.

So the answer is 1-0.9778 = 0.022