Question

1. 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. [10 Marksl a. What is the probability that a randomly selected North American adult watches television for more than 7 hours per day? 14 Marks] 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? [4 Marks] 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? 12 Marks] 2. Because many passengers who make reservations do not show up, airlines often overbook flights (sell more tickets than there are seats). A Boeing 767-400ER holds 245 passengers. If the airline believes the rate of passenger no-shows is 5% and sells 255 tickets, is it likely that it won't have enough seats and someone will get bumped? [10 Marks Use Normal approximation to determine the binomial probability of at least 246 passengers showing up. [9 Marksl a. b. Should the airline change the number of tickets it sells for the flight? Explain. [1 Markl

Answer 1

Here we have $\mu$ = 6, $\sigma$ = 1.5

a. Here we need to find

p ( x > 7 ) = $p\left ( \frac{x-\mu }{\sigma }>\frac{7-6}{1.5} \right )$

= p ( Z > 0.67 )

= 1- p ( z $\leq$ 0.67 )

= 1 - 0.7486

= 0.2514

b) Here n = 5 .

p ( $\overline{x}$ > 7 ) = $p\left ( \frac{x-\mu }{\sigma/\sqrt{n} }>\frac{7-6}{1.5/\sqrt{5}} \right )$

= p ( Z > 1.49 )

= 1- p ( z $\leq$ 1.49 )

= 1 - 0.9319

= 0.0681

c. Here n = 5 and p ( x > 7) = 0.2514

So the required probability is given by

0.2514*0.2514*0.2514*0.2514*0.2514 = 0.2514⁵ = 0.0010