Question

please answer these questions

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. The airline believes the rate of passenger not showing up is 5% and sells 255 tickets. [10 Marks] a. Use Normal approximation to determine the binomial probability of at least 246 passengers showing up. [9 Marks] b. Should the airline change the number of tickets it sells for the flight? Explain. [1 Mark] 3. Suppose at your university, some administrators believe that the proportion of students preferring to take classes at night exceeds 0.30. The president is skeptical and so has an assistant take a simple random sample of 200 students. Of these, 66 indicate that they prefer night classes. What is the probability of finding a sample proportion equal to or greater than that found if the president's skepticism is justified? [10 Marks

Answer 1

a.

Number of tickets sold, n = 255

Probability that a passenger show up = 1 - 0.05 = 0.95

Using Normal approximation, the number of passengers showing up will follow Normal distribution with mean = np = 255 * 0.95 = 242.25 and standard deviation = = 3.48

Vnp(1 - p) = 255 * 0.95 * (1 -0.95)

Probability that at least 246 showing up = P(X $\ge$ 246) = P(X > 245.5) (Using Continuity correction)

= P[Z > (245.5 - 242.25) / 3.48]

= P[Z > 0.9339]

= 0.1752

b.

Since the probability that at least 246 showing up is not very low, (not less than 0.05), the airline should change the number of tickets it sells.

3.

Sample proportion, $\hat{p}$ = 66/200 = 0.33

Hypothesized proportion, p = 0.30

If president's skepticism is justified, the standard error of sample proportion is,

SE = = 0.0324

p(1 - p)/n= 0.3* (1 -0.3)/200

Probability of finding a sample proportion equal to or greater than that found if the president's skepticism is justified

= P(p $\ge$ 0.33 | p = 0.30)

= P[Z $\ge$ (0.33 - 0.30) / 0.0324]

= P[Z $\ge$ 0.9259]

= 0.1772