Respuesta :
Answer:
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.
Step-by-step 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 [tex]\bar X[/tex] = sample mean replacement time
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean replacement time = 3.8 years
[tex]\sigma[/tex] = 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([tex]\bar X[/tex] [tex]\leq[/tex] 3.6 years)
P([tex]\bar X[/tex] [tex]\leq[/tex] 3.6 years) = P( [tex]\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }[/tex] [tex]\leq[/tex] [tex]\frac{3.6-3.8}{\frac{0.4}{\sqrt{44} } }} }[/tex] ) = P(Z [tex]\leq[/tex] -3.32) = 1 - P(Z < 3.32)
= 1 - 0.99955 = 0.0005 or 0.05%
So, in the z table the P(Z [tex]\leq[/tex] 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.