Question

Assume that the amount of time an Internal Revenue Service examiner is supposed to spend reviewing a randomly selected return is 55.1 minutes, with a standard deviation of about 17.6 minutes. If a sample of 35 such reviews is selected, what is the probability that the average review time ismore than an hour?

Answer 1

`P(\bar X >60)=P(Z>(60-55.1)/((17.6)/(√(35)))=1.647)`

And using the complment rule a calculator, excel or the normal standard table we have that:

`P(Z>1.647)=1-P(Z<1.647) = 1-0.9502 =0.0498`

Explanation:

Previous concepts

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".

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".

Solution to the problem

Let X the random variable who represent the amunt of time an internal revenue service examiner i supposed to spend reviewing a selected return, and we know the following properties:

`\mu = 55.1, \sigma =17.6`

We select a sample of n =35>30 so then we can use the central limit theorem.

From the central limit theorem we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

We can find the probability required using the z score formula:

`P(\bar X >60)=P(Z>(60-55.1)/((17.6)/(√(35)))=1.647)`

And using the complment rule a calculator, excel or the normal standard table we have that:

`P(Z>1.647)=1-P(Z<1.647) = 1-0.9502 =0.0498`