The manager of a computer retail 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 3.4 years and a standard deviation of 0.4 years. He then randomly selects records on 34 laptops sold in the past and finds that the mean replacement time is 3.2 years. Assuming that the laptop replacment times have a mean of 3.4 years and a standard deviation of 0.4 years, find the probability that 34 randomly selected laptops will have a mean replacment time of 3.2 years or less.

The probability that 34 randomly selected laptops would have a mean replacement time of 3.2 years or less is 0.0018.

Step-by-step explanation:

Test statistic (z) = (sample mean - population mean) ÷ (population sd/√n)

sample mean = 3.2 years

population mean = 3.4 years

population sd = 0.4 year

n = 34

z = (3.2 - 3.4) ÷ (0.4/√34) = -0.2 ÷ 0.0686 = -2.92

The cumulative area of the test statistic is subtracted from 1 to obtain the probability that 34 randomly selected laptops would have a mean replacement time of 3.2 years or less. The cumulative area is 0.9982. Therefore, the probability is (1 - 0.9982 = 0.0018)

Given Information:  

Population mean = μ = 3.4

Sample mean = x = 3.2

Population standard deviation = σ = 0.4

Number of samples = n = 34

Required Information:  

P(x < 3.2) = ?

P(x < 3.2) = 0.00175

Explanation:

Sample standard deviation is given by

σ = σ√n

σ = 0.4√34

σ = 0.06859

We will find the z-score corresponding to the value of 3.2  

P(x < 3.2) = (x - μ)/ σ

Where σ is the sample standard deviation

P(x < 3.2) = (3.2 - 3.4)/ 0.06859

P(x < 3.2) = -0.2/0.06859

P(x < 3.2) = -2.92

The z-score of x < -2.92 from z-table is

P(z < -2.92) = 0.00175

Therefore, the probability that randomly selected laptops will have a mean replacement time of 3.2 years or less is 0.00175

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step-by-step explanation:

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