Question

The cost of 5 gallons of ice cream has a variance of 49 with a mean of 37 dollars during the summer. What is the probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected? Round your answer to four decimal places.

Answer 1

Answer:

0.0119 = 1.19% probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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

In a set with mean $\mu$ and standard deviation(which is the square root of the variance) $\sigma$, 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 37, \sigma = \sqrt{49} = 7, n = 77, s = \frac{7}{\sqrt{77}} = 0.7977$

What is the probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected?

This is the pvalue of Z when X = 35.2.

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

By the Central Limit Theorem

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

$Z = \frac{35.2 - 37}{0.7977}$

$Z = -2.26$

$Z = -2.26$ has a pvalue of 0.0119

0.0119 = 1.19% probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected