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 0.6 years. He then randomly selects records on 16 laptops sold in the past.

What is the distribution of ¯xx¯? ¯xx¯ ~ N( )

What is the probability that one randomly selected laptop is replaced less than 3.7 years?

For 16 laptops, find the probability that the average replacement time is less than 3.7 year.

For part d), is the assumption of normal necessary? No or Yes

Answer 1

Solution :

Given that ,

mean = = 3.8

standard deviation = = 0.6

a.

X N (3.8 , 0.6)

b.

n = 16

= 3.8

= / n = 0.6/ 16 = 0.15

N (3.8 , 0.15)

c.

P(x < 3.7) = P[(x - ) / < (3.7 - 3.8) / 0.6]

= P(z < -0.17)

= 0.4325

Probability = 0.4325

d.

P( < 3.7) = P(( - ) / < (3.7 - 3.8) / 0.15)

= P(z < -0.67)

= 0.2514

Probability = 0.2514

e.

yes