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 3.8 years and a standard deviation of 0.4 years. He then randomly selects records on 50 laptops sold in the past and finds that the mean replacement time is 3.7 years.

Assuming that the laptop replacement times have a mean of 3.8 years and a standard deviation of 0.4 years, find the probability that 50 randomly selected laptops will have a mean replacement time of 3.7 years or less.
P(M < 3.7 years) =
Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Based on the result above, does it appear that the computer store has been given laptops of lower than average quality?

Yes. The probability of this data is unlikely to have occurred by chance alone.

No. The probability of obtaining this data is high enough to have been a chance occurrence.

Answer 1

Solution

Given population mean laptop replacement times µ = 3.8

Population standard deviation σ = 0.4

Let M be a random variable for the mean replacement time of a sample of 50 laptops

Using Sampling Distribution for the mean as given by Central limit Theorem:

µ_M = µ = 3.8

σ_M = σ/√n = 0.4/√50 = 0.0565685

M ~ N (3.8, 0.05656²)

P ( M < 3.7 years)

Caulculating z value:

z = (m - µ_M)/ σ_M

= (3.7 – 3.8)/ 0.0565685

   = -1.767766

Using z table to find the area to the left of z = -1.767766

P ( M < 3.7 years) = 0.03855

Based on the results, it cannot be concluded that the computer store has been given laptops of lower than average quality as the sample mean (3.7) lies within 2 standard deviations of the mean(3.8) of the sample means which is not an unusual event.

Therefore,

No. The probability of obtaining this data is high enough to have been a chance occurrence.