Question

Installation of a certain hardware device takes random time. The installation time is normally distributed, with a standard deviation of 5 minutes. The mean installation time, however, is unknown. 3.1. [5 pts] A computer technician installs this hardware on 64 different computers, with a measured average installation time of 42 minutes. Compute a 95% confidence interval for the population mean installation time. 3.2. [5 pts] Suppose that the true population mean installation time is 40 minutes. A technician installs the hardware on your PC. What is the probability that the installation time will be within the interval computed in 3.1

Answer 1

According to the information given, we have that:

1. Using the z-distribution, the 95% confidence interval for the population mean installation time, in minutes, is (40.775, 43.225).

2. Using the normal distribution and the central limit theorem, it is found that there is a 0.1075 = 10.75% probability that the installation time will be within the interval computed.

Item 1:

We are given the population standard deviation, which is why the z-distribution is used to solve this question.

The information given is:

• Sample mean of $\overline{x} = 42$.
• Population standard deviation of $\sigma = 5$.
• Samples size of 64, hence $n = 64$.

The confidence interval is:

$\overline{x} \pm M$

The margin of error is given by:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

• z is the critical value.
• $\sigma$ is the population standard deviation.
• n is the sample size.

The first step is finding the critical value, which is z with a p-value of $\frac{1 + \alpha}{2}$, in which $\alpha$ is the confidence level.

In this problem, $\alpha = 0.95$, thus, z with a p-value of $\frac{1 + 0.95}{2} = 0.975$, which means that it is z = 1.96.

Thus:

$M = 1.96\frac{5}{\sqrt{64}} = 1.225$

$\overline{x} - M = 42 - 1.225 = 40.775$

$\overline{x} + M = 42 + 1.225 = 43.225$

The 95% confidence interval for the population mean installation time, in minutes, is (40.775, 43.225).

Item 2:

First, the normal distribution and the central limit theorem are introduced.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• By the Central Limit Theorem, for sampling distributions of samples means of size n, the standard deviation is $s = \frac{\sigma}{\sqrt{n}}$.

For this problem, we have that $\mu = 40, s = \frac{5}{\sqrt{64}} = 0.625$

The probability is the p-value of Z when X = 43.225 subtracted by the p-value of Z when X = 40.775.

X = 43.225:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{43.225 - 40}{0.625}$

$Z = 5.16$

$Z = 5.16$ has a p-value of 1.

X = 40.775:

$Z = \frac{X - \mu}{s}$

$Z = \frac{40.775 - 40}{0.625}$

$Z = 1.24$

$Z = 1.24$ has a p-value of 0.8925.

1 - 0.8925 = 0.1075.

0.1075 = 10.75% probability that the installation time will be within the interval computed.

A similar problem is given at https://brainly.com/question/24663213