Question

6) Supplementary Exercise 5.51

A consumer advocate claims that 80 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.

(a)

Suppose that the advocate's claim is true, and suppose that a random sample of five cable subscribers is selected. Assuming independence, use an appropriate formula to compute the probability that four or more subscribers in the sample are not satisfied with their service. (Do not round intermediate calculations. Round final answer of p to 1 decimal place. Round other final answers to 4 decimal places.)

The answer for 6(a) is P( Xâ¥4) = P ( x = 4) + P (x = 5) = 5/4 * 0.84 * 0.21 + 5/5 * 0.85 * 0.20 = 0.737

(b)

Suppose that the advocate's claim is true, and suppose that a random sample of 25 cable subscribers is selected. Assuming independence, find each of the following: (Do not round intermediate calculations. Round final answer of p to 1 decimal place. Round other final answers to 4 decimal places.)

1.

The probability that 15 or fewer subscribers in the sample are not satisfied with their service.

The answer for 6(b)1 is P(Yâ¤15) = 1 - P( Y > 20) - X20, i = 16 P ( Y = i) = 1- 0.421 - 0.562 = 0.017

2.

The probability that more than 20 subscribers in the sample are not satisfied with their service.

The answer for 6(b)2 is

3.

The probability that between 20 and 24 (inclusive) subscribers in the sample are not satisfied with their service.

The answer for 6(b)3 is P(20 > Y < 24) = 1 - 0.421 - 0.1867 - 0.1358 - 0.0708 - 0.0236 = 0.1621

4.

The probability that exactly 24 subscribers in the sample are not satisfied with their service.

The answer for 6(b)4 is P( Y = 24) = 0.0236

(c)

Suppose that when we survey 25 randomly selected cable television subscribers, we find that 15 are actually not satisfied with their service. Using a probability you found in this exercise as the basis for your answer, do you believe the consumer advocate's claim? Explain. (Round your answer to 4 decimal places.)

Answer 1

`P(X \le 4) = 0.7373`

`P(x \le 15) = 0.0173`

`P(x > 20) = 0.4207`

`P(20\ge x \le 24)= 0.6129`

`P(x = 24) = 0.0236`

`P(x = 15) = 1.18\%`

Step-by-step explanation:

Given

`p = 80\% = 0.8`

The question illustrates binomial distribution and will be solved using:

`P(X = x) = ^nC_xp^x(1 - p)^(n-x)`

Solving (a):

Given

`n =5`

Required

`P(X\ge 4)`

This is calculated using

`P(X \le 4) = P(x = 4) +P(x=5)`

This gives:

`P(X \le 4) = ^5C_4 * (0.8)^4*(1 - 0.8)^(5-4) + ^5C_5*0.8^5*(1 - 0.8)^(5-5)`

`P(X \le 4) = 5 * (0.8)^4*(0.2)^1 + 1*0.8^5*(0.2)^0`

`P(X \le 4) = 0.4096 + 0.32768`

`P(X \le 4) = 0.73728`

`P(X \le 4) = 0.7373` --- approximated

Solving (b):

Given

`n =25`

i)

Required

`P(X\le 15)`

This is calculated as:

`P(X\le 15) = 1 - P(x>15)` --- Complement rule

`P(x>15) = P(x=16) + 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)`

`P(x > 15) = {25}^C_(16) * p^(16)*(1-p)^(25-16) +{25}^C_(17) * p^(17)*(1-p)^(25-17) +{25}^C_(18) * p^(18)*(1-p)^(25-18) +{25}^C_(19) * p^(19)*(1-p)^(25-19) +{25}^C_(20) * p^(20)*(1-p)^(25-20) +{25}^C_(21) * p^(21)*(1-p)^(25-21) +{25}^C_(22) * p^(22)*(1-p)^(25-22) +{25}^C_(23) * p^(23)*(1-p)^(25-23) +{25}^C_(24) * p^(24)*(1-p)^(25-24) +{25}^C_(25) * p^(25)*(1-p)^(25-25)`

`P(x > 15) = 2042975 * 0.8^(16)*0.2^9 +1081575* 0.8^(17)*0.2^8 +480700 * 0.8^(18)*0.2^7 +177100 * 0.8^(19)*0.2^6 +53130 * 0.8^(20)*0.2^5 +12650 * 0.8^(21)*0.2^4 +2300 * 0.8^(22)*0.2^3 +300 * 0.8^(23)*0.2^2 +25* 0.8^(24)*0.2^1 +1 * 0.8^(25)*0.2^0`

`P(x > 15) = 0.98266813045`

So:

`P(X\le 15) = 1 - P(x>15)`

`P(x \le 15) = 1 - 0.98266813045`

`P(x \le 15) = 0.01733186955`

`P(x \le 15) = 0.0173`

ii)

`P(x>20)`

This is calculated as:

`P(x>20) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)`

`P(x > 20) = 12650 * 0.8^(21)*0.2^4 +2300 * 0.8^(22)*0.2^3 +300 * 0.8^(23)*0.2^2 +25* 0.8^(24)*0.2^1 +1 * 0.8^(25)*0.2^0`

`P(x > 20) = 0.42067430925`

`P(x > 20) = 0.4207`

iii)

`P(20\ge x \le 24)`

This is calculated as:

`P(20\ge x \le 24) = P(x = 20) + P(x = 21) + P(x = 22) + P(x =23) + P(x = 24)`

`P(20\ge x \le 24)= 53130 * 0.8^(20)*0.2^5 +12650 * 0.8^(21)*0.2^4 +2300 * 0.8^(22)*0.2^3 +300 * 0.8^(23)*0.2^2 +25* 0.8^(24)*0.2^1`

`P(20\ge x \le 24)= 0.61291151859`

`P(20\ge x \le 24)= 0.6129`

iv)

`P(x = 24)`

This is calculated as:

`P(x = 24) = 25* 0.8^(24)*0.2^1`

`P(x = 24) = 0.0236`

Solving (c):

`P(x = 15)`

This is calculated as:

`P(x = 15) = {25}^C_(15) * 0.8^(15) * 0.2^(10)`

`P(x = 15) = 3268760 * 0.8^(15) * 0.2^(10)`

`P(x = 15) = 0.01177694905`

`P(x = 15) = 0.0118`

Express as percentage

`P(x = 15) = 1.18\%`

The calculated probability (1.18%) is way less than the advocate's claim.

Hence, we do not believe the claim.