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 is the probability that a randomly selected
North American adult watches television for
more than 7 hours per day?
b. What is the probability that the average time
watching television by a random sample of five
North American adults is more than 7 hours?
c. What is the probability that in a random sample
of five North American 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

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