Question

An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic random sampling of 100 filings. These 100 filings showed an average preparation time of 90 minutes with a standard deviation of 140 minutes.
A) What is the standard error of the mean?
B) What is the probability that the mean completion time will be more than 120 minutes?

Answer 1

Answer:

a)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 error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

b) We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

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

Part a

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 error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

Part b

We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$