Question

A poll reported that 65 % 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 is recorded. Complete parts (a) through (e) below.

(c) Find and interpret the probability that at least 21 of them are satisfied with the airlines.

Answer #1

Given P = 0.65, n = 25

If p = 0.65, q = 1-p = 0.35

Binomial probability

P(X = r) = nCr * p^r * q^n-r

The probability that at least 21 of them are satisfied with the airlines

P(X ≥ 21) = P(X = 21)+P(X = 22)+P(X = 23)+P(X = 24)+P(X = 25)

P(X = 21) = ^{25}C_{21} * 0.65^21 * 0.35^4 =
25!/21!(25−21)! * 0.65^21 * 0.35^4 = 0.0224

P(X = 22) = ^{25}C_{22} * 0.65^22 * 0.35^3 =
25!/22!(25−22)! * 0.65^22 * 0.35^3 = 0.0076

P(X = 23) = ^{25}C_{23} * 0.65^23 * 0.35^2 =
25!/23!(25−23)! * 0.65^23 * 0.35^2 = 0.0018

P(X = 24) = ^{25}C_{24} * 0.65^24 * 0.35^1 =
25!/24!(25−24)! * 0.65^24 * 0.35^1 = 0.0003

P(X = 25) = ^{25}C_{25} * 0.65^25 * 0.35^0 =
25!/25!(25−25)! * 0.65^25 * 0.35^0 = 2.10297E-5

P(X ≥ 21) = 0.0224 + 0.0076 + 0.0018 + 0.0003 + 2.10297E-5 = 0.0321

the probability that at least 21 of them are satisfied with the airlines is 0.0321

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