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 44 laptops sold in the past and finds that the mean replacement time is 3.6 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 44 randomly selected laptops will have a mean replacement time of 3.6 years or less.

P(¯¯¯X≤3.6 years)P(X¯≤3.6 years) = Round to 4 decimal places.

NOTE: Answers obtained using exact z-scores or z-scores rounded to 2 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?

No. The probability of obtaining this data is greater than 5%, high enough to have been a chance occurrence.

Yes. The probability of obtaining this data is less than 5%, so it is unlikely to have occurred by chance alone.

Answer 1

Probability that the 44 randomly selected laptops will have a mean replacement time of 3.6 years or less is 0.0092.

Yes. The probability of obtaining this data is less than 5%, so it is unlikely to have occurred by chance alone.

Explanation:

We are given that the 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 44 laptops sold in the past and finds that the mean replacement time is 3.6 years.

Let
`\bar X` = sample mean replacement time

The z-score probability distribution for sample mean is given by;

Z =
`( \bar X-\mu)/((\sigma)/(√(n) ) )} }` ~ N(0,1)

where,
`\mu` = population mean replacement time = 3.8 years

`\sigma` = standard deviation = 0.4 years

n = sample of laptops = 44

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the 44 randomly selected laptops will have a mean replacement time of 3.6 years or less is given by = P(
`\bar X`
`\leq` 3.6 years)

P(
`\bar X`
`\leq` 3.6 years) = P(
`( \bar X-\mu)/((\sigma)/(√(n) ) )} }`
`\leq`
`(3.6-3.8)/((0.4)/(√(44) ) )} }` ) = P(Z
`\leq` -3.32) = 1 - P(Z < 3.32)

= 1 - 0.99955 = 0.0005 or 0.05%

So, in the z table the P(Z
`\leq` x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 3.32 in the z table which has an area of 0.99955.

Hence, the required probability is 0.0005 or 0.05%.

Now, based on the result above; Yes, the computer store has been given laptops of lower than average quality because the probability of obtaining this data is less than 5%, so it is unlikely to have occurred by chance alone.