Question

For a certain type of computers, the length of time between charges of the battery is normally distributed with a mean of 42 hours and a standard deviation of 18 hours. a. If 20 computers are randomly selected, what is the probability that the mean charging time of their batteries is more than 45 hours? b. If 38 computers are randomly selected, what is the probability that the mean charging time of their batteries is between 37 and 40 hours? c. If 16 computers are randomly selected, what is the probability that the mean charging time of their batteries is less than 50 hours? Will this be unusual?

Answer 1

Solution:- Given that mean = 42, sd = 18

a. for n = 20,

P(X > 45) = P((X-mean)/(sd/sqrt(n)) > (45-42)/((18/sqrt(20)))
= P(Z > 0.7454)
= 1 − P(Z < 0.7454)
= 1 − 0.7734
= 0.2266

b. for n = 38

P(37 < X < 40) = P((37-42)/(18/sqrt(38)) < Z < (40-42)/(18/sqrt(38)))
= P(-1.7123 < Z < -0.6849)
= P(Z < −0.6849) − P( Z < −1.7123)
= 0.2483 - 0.0436
= 0.2047

c. for n = 16

P(X < 50) = P(Z < (50-42)/(18/sqrt(16)))
= P(Z < 1.7778)
= 0.9625