Answer to Question #153626 in Statistics and Probability for nitin sher

Question

Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times spent on entertainment are normally distributed and the standard deviation for the times is half an hour.

(a) Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.

(b) Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.

Answer 1

Let X be a random variable denoting the number of hours.

X~N(2,0.5²)

a) P(1.8<X<2.75)

Z=$\frac{x-\mu} {\sigma}$

$P(X<1.8)$

$Z_1=\frac{1.8-2}{0.5}=-0.4$

P(z<-0.4)=0.34458 from the z tables

$P(X<2.75)$

$Z_2=\frac{2.75-2}{0.5}$ =1.5

P(z<1.5)=0.93319 from the z tables

P(-0.4<z<1.5)=0.93319-0.34458

=0.58861

b) P(X<x) =0.25

$\Phi ^{-1}(0.25)=-0.68$

$-0.68=\frac{x-2}{0.5}$

x=1.66 hours