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

Answer:

b. 0.0222

Step-by-step 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 $\frac{\sigma}{\sqrt{n}}$.

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:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 402.7, \sigma = 8.8, n = 40, s = \frac{8.8}{\sqrt{40}} = 1.39$

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

$Z = \frac{X - \mu}{s}$

$Z = \frac{405.5 - 402.7}{1.39}$

$Z = 2.01$

$Z = 2.01$ has a pvalue of 0.9778.

So the answer is 1-0.9778 = 0.022