A consumer advocate claims that 75 percent of cable television subscribers are n

Question

A consumer advocate claims that 75 percent of cable television subscribers are not satisfied with their cable service. In an attempt to justify this claim, a randomly selected sample of cable subscribers will be polled on this issue. Suppose that the advocate's claim is true, and suppose that a random sample of 4 cable subscribers is selected. Assuming independence, use an appropriate formula to compute the probability that 3 or more subscribers in the sample are not satisfied with their service. (Do not round intermediate calculations. Round final answer to p in 2 decimal place. Round other final answers to 4 decimal places.) Binomial, n = 4, p =0.75 Probability=______________ Suppose that the advocate's claim is true, and suppose that a random sample of 20 cable subscribers is selected. Assuming independence, find: (Do not round intermediate calculations. Round final answer to pin 2 decimal place. Round other final answers to 4decimal places.) Binomial, n =________, p=_______________. The probability that 15 or fewer subscribers in the sample are not satisfied with their service. Probability 0.0713 The probability that more than 16 subscribers in the sample are not satisfied with their service. Probability 8506 The probability that between 16 and 18 (inclusive) subscribers in the sample are not satisfied with their service. Probability_____________ The probability that exactly 18 subscribers in the sample are not satisfied with their service. Probability 0.5611

Explanation / Answer

Binomial Distribution
PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
a.
P( X < 3) = P(X=2) + P(X=1) + P(X=0)
= ( 4 2 ) * 0.75^2 * ( 1- 0.75 ) ^2 + ( 4 1 ) * 0.75^1 * ( 1- 0.75 ) ^3 + ( 4 0 ) * 0.75^0 * ( 1- 0.75 ) ^4
= 0.2617
P( X > = 3 ) = 1 - P( X < 3) = 0.7383
b.
n=20, p =.075
1.
P( X < = 15) = P(X=15) + P(X=14) + P(X=13) + P(X=12) + P(X=11) + P(X=10) ... +P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 20 15 ) * 0.75^15 * ( 1- 0.75 ) ^5 + ( 20 14 ) * 0.75^14 * ( 1- 0.75 ) ^6 + ( 20 13 ) * 0.75^13 * ( 1- 0.75 ) ^7 + ( 20 12 ) * 0.75^12 * ( 1- 0.75 ) ^8 + ( 20 11 ) * 0.75^11 * ( 1- 0.75 ) ^9 + ( 20 10 ) * 0.75^10 * ( 1- 0.75 ) ^10 .............+ ( 20 3 ) * 0.75^3 * ( 1- 0.75 ) ^17 + ( 20 2 ) * 0.75^2 * ( 1- 0.75 ) ^18 + ( 20 1 ) * 0.75^1 * ( 1- 0.75 ) ^19 + ( 20 0 ) * 0.75^0 * ( 1- 0.75 ) ^20
= 0.5852
2.
P( X > = 16 ) = 1 - P( X < 16) = 0.4148
3.
P( X = 16 ) = ( 20 16 ) * ( 0.75^16) * ( 1 - 0.75 )^4
= 0.1897
P( X = 17 ) = ( 20 17 ) * ( 0.75^17) * ( 1 - 0.75 )^3
= 0.1339
P( X = 18 ) = ( 20 18 ) * ( 0.75^18) * ( 1 - 0.75 )^2
= 0.0669

P(16<=X<=18) = 0.1897 + 0.1339 + 0.0669 = 0.3905

4.
P( X = 18 ) = ( 20 18 ) * ( 0.75^18) * ( 1 - 0.75 )^2
= 0.0669

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