Answer to Question #330659 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. (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

Let $X=$ the amount of time spent by North American adults watching television per day: $X\sim N(\mu, \sigma^2).$

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

a) Given $\mu=6\ hours, \sigma=1.5 \ hours$

$$P(X>7)=1-P(X\leq7)$$ $$=1-P(Z\leq\dfrac{7-6}{1.5})\approx 1-P(Z<0.666667)$$ $$\approx1-0.747507\approx0.252493$$

b) Given $\mu=6\ hours, \sigma=1.5 \ hours, n=5$

$$X\sim N(\mu, \sigma^2/n), Z=\dfrac{X-\mu}{\sigma/\sqrt{n}}\sim N(0, 1)$$ $$P(X>7)=1-P(X\leq7)$$ $$=1-P(Z\leq\dfrac{7-6}{1.5/\sqrt{5}})\approx 1-P(Z<1.490712)$$ $$\approx1-0.931981\approx0.068019$$

b) Given $\mu=6\ hours, \sigma=1.5 \ hours, n=30$

$$X\sim N(\mu, \sigma^2/n), Z=\dfrac{X-\mu}{\sigma/\sqrt{n}}\sim N(0, 1)$$ $$P(X>7)=1-P(X\leq7)$$ $$=1-P(Z\leq\dfrac{7-6}{1.5/\sqrt{5}})\approx 1-P(Z<1.490712)$$ $$\approx1-0.931981\approx0.068019$$

c) The likelihood that five out of five adults watch TV more than 7 hours a day is the first probability raised to the fifth power

$$(0.252493)^5 \approx0.001026$$