Answer to Question #332550 in Statistics and Probability for lovely

Question

Replacement times of TV sets are reported to follow a normal distribution having a mean

of 8.5 years with standard deviation of 1.2 years.

a. If 30 TV sets are selected at random, what is the probability that the mean

replacement time is less than 8 years?

b. If 20 TV sets are selected, what is the probability that the mean replacement

time is longer than 7.8 years?

c. If 25 TV sets are randomly selected, what is the probability that the

replacement time is between 8.4 years and 9 years?

Answer 1

a. $P(X<8)=P(Z<\frac{8-8.5}{1.2/\sqrt{30}})=P(Z<-2.28)=0.0113$

b.

$P(X>7.8)=1-P(Z<\frac{7.8-8.5}{1.2/\sqrt{20}})=1-P(Z<-2.61)=1-0.00453=0.99547$

c.$P(8.4<X<9)=P(\frac{8.4-8.5}{1.2/\sqrt{25}}<Z<\frac{9-8.5}{1.2/\sqrt{25}})=P(-0.42<Z<2.08)=0.9812-0.33724=0.64396$