Question

# A poll reported that 65 ​% of adults were satisfied with the job the major airlines...

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.

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) = 25C21 * 0.65^21 * 0.35^4 = 25!/21!(25−21)! * 0.65^21 * 0.35^4 = 0.0224

P(X = 22) = 25C22 * 0.65^22 * 0.35^3 = 25!/22!(25−22)! * 0.65^22 * 0.35^3 = 0.0076

P(X = 23) = 25C23 * 0.65^23 * 0.35^2 = 25!/23!(25−23)! * 0.65^23 * 0.35^2 = 0.0018

P(X = 24) = 25C24 * 0.65^24 * 0.35^1 = 25!/24!(25−24)! * 0.65^24 * 0.35^1 = 0.0003

P(X = 25) = 25C25 * 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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