Answer to Question #97502 in Statistics and Probability for mahsa

Question

A survey was conducted to measure the number of hours per week adults spend on home computers. In the survey, the number of hours were normally distributed, with a mean of 7 hours and a standard deviation of 1 hour. Find the probability that the hours spent on the home computer by the participant are less than 4.5 hours per week.

Answer 1

Let $X=$ the number of hours per week adults spend on home computers: $X\sim N(\mu, \sigma^2)$

Then

$$Z={X-\mu \over \sigma}\sim N(0,1)$$

Given that $\mu=7$ hours and $\sigma=1$ hour.

$$P(X<4.5)=P(Z<{4.5-7 \over 1})=P(Z<-2.5)\approx0.0062$$

The probability that the hours spent on the home computer by the participant are less than 4.5 hours per week is $0.0062$ .