Question

The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than average quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 4.1 years and a standard deviation of 0.4 years. He then randomly selects records on 38 laptops sold in the past. Round the answers of following questions to 4 decimal places.

Answer 1

Given:

Population mean = 4.1 years

Population standard deviation = 0.4

Sample size = 38

Find:

b. distribution of sample mean

c. probability of selecting a laptop that has been replaced less than 4.2 years.

d. probability of having an average replacement of fewer than 4.2 years.

e. Is the assumption of normality necessary for part d

Solution:

To determine the distribution of the sample mean, we need to calculate the standard deviation of the sample. The formula is:

$s=\frac{\sigma}{\sqrt{n}}$

where σ is the population standard deviation and n = sample size.

Since we already have this information given in the problem, let's plug it into the formula above and solve for s.

$s=\frac{0.4}{\sqrt{38}}\Rightarrow s=\frac{0.4}{6.164414}\Rightarrow s=0.0649$

The sample standard deviation of the mean is 0.0649.

b. Hence, the distribution of the sample mean is N (4.1, 0.0649).

c. To get the probability of selecting a laptop that has been replaced for less than 4.2 years, we need to convert the random variable x which is 4.2 years to a z-score. We used the formula below:

$z=\frac{x-\mu}{\sigma}$

where x = 4.2, μ = 4.1, and σ = 0.4. Let's plug this into the formula above and solve for z.

$z=\frac{4.2-4.1}{0.4}\Rightarrow z=\frac{0.1}{0.4}\Rightarrow z=0.25$

The z-score equivalent of 4.2 years is 0.25. Looking at the standard normal distribution table, the area covered under the normal curve less than z = 0.25 is:

$P(z<0.25)=0.5987$

Therefore, the probability of selecting a laptop that has been replaced for less than 4.2 years is 0.5987.

Now, for the average replacement of fewer than 4.2 years, we will have to convert again 4.2 to a z-score but this time, we will use a different formula because the given 4.2 years is not a random variable but a random average. Formula is:

$z=\frac{\bar{x}-\mu}{s}$

where bar x = 4.2, μ = 4.1, and s = 0.0649. Let's plug this into the formula above and solve for z.

$z=\frac{4.2-4.1}{0.0649}\Rightarrow z=\frac{0.1}{0.0649}\Rightarrow z=1.54$

Looking at the standard normal distribution table, the area covered under the normal curve of less than z = 1.54 is:

$P(z<1.54)=0.9382$

Therefore, the probability of having an average replacement of fewer than 4.2 years is 0.9382.

e. No, the assumption of normality is not necessary to determine the probability in part d.