When an opinion poll calls residential telephone numbers at random, only 17% of

Question

When an opinion poll calls residential telephone numbers at random, only 17% of the calls reach a live person. You watch the random dialing machine make 18 calls. (Round your answers to four decimal places.) (a) What is the probability that exactly 3 calls reach a person? (b) What is the probability that at most 3 calls reach a person? (c) What is the probability that at least 3 calls reach a person? (d) What is the probability that fewer than 3 calls reach a person? (e) What is the probability that more than 3 calls reach a person?

Explanation / Answer

Solution:-

  Let N be the number of live persons contacted among the 18 calls observed. Then N has the binomial distribution with n = 18 and p = 0.17.

(a) P(N = 3) = (18C3)(0.17)3(0.83)15

   = 816 * 0.00030027429

= 0.245.

(b) P(N 3) = P(N = 0) + ··· + P( N = 3)

= (18C0)(0.17)0(0.83)18 + (18C1)(0.17)1(0.83)17 + (18C2)(0.17)2(0.83)16 + (18C3)(0.17)3(0.83)15

= 1*1*0.03494665903 + 18*0.17*0.04210440848 + 153*0.00146604506 + 816*0.00030027429

= 0.6331

(c) P(N 3) = P(N = 3) + ··· + P(N = 15)

= 1 - (P(N=0) + P(N=1) + P(N=2))

= 1 - { (18C0)(0.17)0(0.83)18 + (18C1)(0.17)1(0.83)17 + (18C2)(0.17)2(0.83)16 }

= 1 - (1*1*0.03494665903 + 18*0.17*0.04210440848 + 153*0.00146604506)

= 0.6119

(d) P(N < 3) = P(N = 0) + ··· + P(N = 2)

= (P(N=0) + P(N=1) + P(N=2))

= (18C0)(0.17)0(0.83)18 + (18C1)(0.17)1(0.83)17 + (18C2)(0.17)2(0.83)16

   =  (1*1*0.03494665903 + 18*0.17*0.04210440848 + 153*0.00146604506)

= 0.3881

(e) P(N > 3) = P(N = 4) + ··· + P(N = 15)

= 1 - {P(N < 3)}

= 1 - 0.6331

= 0.3669

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