Answer to Question #330652 in Statistics and Probability for mau

Question

The amount of time spent by North American adults watching television per day is normally distributed with a mean of 6 hours and a standard deviation of 1.5 hours. (30%) (a) What proportion of the population watches television for more than 7 hours per day? (b) What is the probability that the average number of hours spent watching television by a random sample of five adults is more than 7 hours? (c) What is the probability that in a random sample of five adults all watch television for more than 7 hours per day? 

Answer 1

1) X~$N(6,1.5^2)$

$P(X>7)=P(N(6,1.5^2)>7)=P(6+1.5Z>7)=P(Z>0.67)=0.25143$

2) Y~$N(6,({\frac {1.5} {\sqrt{5}}})^2)$

$P(Y>7)=P(N(6,({\frac {1.5} {\sqrt{5}}})^2)>7)=P(6+0.67Z>7)=P(Z>1.5)=0.06681$

3) $P=(P(X>7))^5=0.25143^5=0.001$