Bones Brothers & Associates prepare individual tax returns. Over prior years, Bones Brothers have maintained careful records

Question

2 months ago

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Bones Brothers & Associates prepare individual tax returns. Over prior years, Bones Brothers have maintained careful records

regarding the time to prepare a return. The mean time to prepare a return is 90 minutes, and the population standard deviation of this distribution is 14 minutes. Suppose 49 returns from this year are selected and analyzed regarding the preparation time. What is the standard deviation of the sample mean? Select one: a. 14 minutes b. 2 minutes c. .28 minutes d. 98 minutes

Mathematics

1 answer:

Leona [12.6K] · Answer 1 · 2026-03-25T17:33:45+03:00

Response:

Regarding this scenario, we have the following details about the timeframe to organize a return

$\mu =90, \sigma =14$

And we select a sample size =49>30, and we aim to find the standard deviation for the sample mean. Based on the central limit theorem, we understand that the sample mean distribution $\bar X$ is represented by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

The standard deviation would then be:

$\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2$

Therefore, the most accurate response would be

b. 2 minutes

Step-by-step explanation:

Prior concepts

Normal distribution, defines a probability distribution that is symmetric around the mean, showing that occurrences near the mean are more frequent than those that are further away.

The central limit theorem states that "when we have a population with mean μ and standard deviation σ and draw sufficiently large random samples from the population with replacement, then the distribution of the sample means will resemble a normal distribution. This applies regardless of whether the original population is normal or skewed, given that the sample size is large enough".

Solution to the problem

For this case, we have the following information about the time to prepare a return

$\mu =90, \sigma =14$

A sample size =49>30 also indicates that we are focused on determining the standard deviation for the sample mean. Based on the central limit theorem, we recognize that the sample mean distribution $\bar X$ is described by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And subsequently, the standard deviation will be:

$\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2$

Hence, the most accurate answer would be

b. 2 minutes