Question

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

Answer 1

Answer:

For this case we have the following info related to the time to prepare a return

$\mu =90 , \sigma =14$

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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

And the standard deviation would be:

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

And the best answer would be

b. 2 minutes

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

For this case we have the following info related to the time to prepare a return

$\mu =90 , \sigma =14$

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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

And the standard deviation would be:

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

And the best answer would be

b. 2 minutes