Question

Great Value Airlines routinely surveys its passengers to determine their satisfaction with their flight. Historically, 95 percent of passengers, when surveyed by email within a week of flying, express satisfaction with their recent flight. Assuming that nothing special happened on a given flight, what is the probability that at most 10 of the 150 passengers on that flight express the opinion that the flight was unsatisfactory?

• A. 0.9612

• B. 0.7677

• C. 0.4328

• D. 0.8695

Answer 1

n = 150

p = Probability of unsatisfactory = 0.05

q = 1 - p = 0.95

= np = 150 X 0.05 = 7.5

To find P(X10):

Applying Continuity Correction, we get:

To find P(X<10.5):

Z = (10.5 - 7.5)/2.6693

= 1.1239

By Technology, cumulative area = 0.8695

So,

P(X10)= 0.8695

So,

Correct option:

D.     0.8695