Question

The cost of 5 gallons of ice cream has a variance of 49 with a mean of 31 dollars during the summer. What is the probability that the sample mean would be less than 32.3 dollars if a sample of 31 5-gallon pails is randomly selected? Round your answer to four decimal places. (show work)

Answer 1

Using the Central Limit Theorem, the standard error of the mean was calculated, followed by the z-score for the given sample mean. Then, the cumulative probability for the z-score was looked up, revealing a probability of approximately 0.8495, indicating a high likelihood that the sample mean cost of 5 gallons of ice cream would be less than $32.3.

Step-by-step explanation:

To find the probability that the sample mean would be less than $32.3 when the mean cost of 5 gallons of ice cream is $31 and the variance is 49 during the summer, we would use the Central Limit Theorem (CLT). The CLT states that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough, regardless of the population's distribution. Since we are dealing with a sample of 31, which is generally considered a large enough sample, the CLT applies.

We need to calculate the standard error (SE) of the mean, which is the standard deviation of the sampling distribution. The formula for standard error is SE = √(variance) / √(sample size). Given the variance is 49, and the sample size is 31, SE = √(49) / √(31) = 7 / √(31) ≈ 1.257.

Next, we calculate the z-score, which is the number of standard errors the sample mean is away from the population mean. The formula is z = (X - μ) / SE, where X is the sample mean, and μ is the population mean. For a sample mean of $32.3, z = (32.3 - 31) / 1.257 ≈ 1.035.

Finally, we look up the cumulative probability for z = 1.035 in the standard normal distribution table or use a statistical software to find the probability. This value corresponds to the probability that we would observe a sample mean less than $32.3.

If we assume a normal distribution, this would give us the probability that the z-score is less than 1.035. Using statistical tables or software, we can find that P(Z < 1.035) ≈ 0.8495. The probability of observing a sample mean less than $32.3 is approximately 0.8495 or 84.95%, so the probability that the sample mean would be less than $32.3 is about 0.8495, rounded to four decimal places.