Question

# A poll reported that 63% of adults were satisfied with the job the major airlines were...

A poll reported that 63% of adults were satisfied with the job the major airlines were doing. Suppose 25 adults are selected at random and the number who are satisfied are recorded. Complete parts A.) through D.)

Part A.) Find and Interpret the probability that exactly 14 of them are satisfied with the Airline.

Part B.) Find and interpret the probability that at least 17 of them are satisfied with the airlines

Part C.) Find and interpret the probability that between 11 and 14 of them, inclusive, are satisfied with airlines.

Part D.) Would it be unusual to find more than 22 who are satisfied with the job the major airlines were doing?

n = 25

P = 0.63

It is a binomial distribution.

P(X = x) = nCx * px * (1 - p)n - x

A) P(X = 14) = 25C14 * (0.63) ^14 * (0.37) ^11 = 0.1230

B) P(X > 17) = P (X = 17) + P(X = 18) + P(X = 19) + P(X = 20) + P(X = 21) + P(X = 22) + P(X = 23) + P(X = 24) + P(X = 25)

= 25C17 * +0.63) ^17 * (0.37) ^8 + 25C18 * (0.63) ^18 * (0.37) ^7 + 25C19 * (0.63) ^19 * (0.37) ^6 + 25C20 * (0.63) ^20 * (0.37) ^5 + 25C21 * (0.63) ^21 * (0.37) ^4 + 25C22 * (0.63) ^22 * (0.37) ^3 + 25C23 * (0.63) ^23 * (0.37) ^2 + 25C24 * (0.63) ^24 * (0.37) ^1 + 25C25 * (0.63) ^25 * (0.37) ^0 = 0.3658

C) P(11 < X < 14) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 25C11 * (0.63) ^11 * (0.37) ^14 + 25C12 * (0.63) ^12 * (0.37) ^13 + 25C13 * (0.63) ^13 * (0.37) ^12 + 25C14 * (0.63) ^14 * (0.37) ^11 = 0.3852

D) P(X > 22) = P(X = 23) + P(X = 24) + P(X = 25)

= 25C23 * (0.63) ^23 * (0.37) ^2 + 25C24 * (0.63) ^24 * (0.37) ^1 + 25C25 * (0.63) ^25 * (0.37) ^0 = 0.0055

As the probability is less than 0.05, so it would be unusual to find more than 22 who are satiated with the job the major airlines we're doing.

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