Question

The cost of 5 gallons of ice cream has a standard deviation of 6 dollars with a mean of 32 dollars during the summer. What is the probability that the sample mean would be greater than 33.3 dollars if a sample of 94 5-gallon pails is randomly selected? Round your answer to four decimal places.

Answer 1

The probability that the sample mean would be greater than 33.3 dollars is 0.0179.

Explanation:

We are given that the cost of 5 gallons of ice cream has a standard deviation of 6 dollars with a mean of 32 dollars during the summer.

A sample of 94 5-gallon pails is randomly selected.

Let
`\bar X` = sample mean cost

The z-score probability distribution for sample mean is given by;

Z =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = population mean cost = 32 dollars

`\sigma` = population standard deviation = 6 dollars

n = sample of 5-gallon pails = 94

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the sample mean would be greater than 33.3 dollars if a sample of 94 5-gallon pails is randomly selected is given by = P(
`\bar X` > 33.3 dollars)

P(
`\bar X` > 33.3 dollars) = P(
`(\bar X-\mu)/((\sigma)/(√(n) ) )` >
`(33.3-32)/((6)/(√(94) ) )` ) = P(Z > 2.10) = 1 - P(Z
`\leq` 2.10)

= 1 - 0.98214 = 0.0179

So, in the z table the P(Z
`\leq` x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.10 in the z table which has an area of 0.98214.

Hence, the probability that the sample mean would be greater than 33.3 dollars is 0.0179.