A consumer advocate claims that 80% 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.

 

Round to four decimal places.

 

(a) Suppose that the advocate's claim is true, and suppose that a random sample of 6 cable subscribers is selected. Calculate the probability that 4 or more subscribers in the sample are not satisfied with their service. 

 

Binomial, n = 
 , p = 
       

Probability  = 
                               
 

(b) Suppose that the advocate's claim is true, and suppose that a random sample of 20 cable subscribers is selected.

 

Binomial, n = 
 , p = 

 

Calculate the following: 

 

1. The probability that 12 or fewer subscribers in the sample are not satisfied with their service.
 

       Probability =  

 

2. The probability that more than 14 subscribers in the sample are not satisfied with their service.


        Probability =  
  


3. The probability that between 14 and 18 (inclusive) subscribers in the sample are not satisfied with their service.


        Probability =  


4. The probability that exactly 18 subscribers in the sample are not satisfied with their service.


        Probability =  


To solve this problem, we will use the **binomial distribution** formula, which is applicable for scenarios where there are two possible outcomes, such as "satisfied" or "not satisfied." The binomial distribution formula is:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where:
- \(n\) is the number of trials (sample size),
- \(k\) is the number of successful trials (in this case, the number of unsatisfied subscribers),
- \(p\) is the probability of success (here, the probability that a subscriber is unsatisfied),
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).

### (a) Probability that 4 or more out of 6 subscribers are not satisfied
Given:
- \(n = 6\) (sample size),
- \(p = 0.8\) (probability that a subscriber is not satisfied).

We need to calculate \(P(X \geq 4)\), i.e., the probability that 4 or more subscribers are not satisfied. This can be calculated by summing the individual binomial probabilities for \(X = 4\), \(X = 5\), and \(X = 6\).

\[
P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)
\]

I'll calculate this.

### (b) Sample of 20 subscribers:
Given:
- \(n = 20\),
- \(p = 0.8\).

#### 1. The probability that 12 or fewer subscribers are not satisfied
We need to calculate \(P(X \leq 12)\), i.e., the cumulative probability of \(X = 0\) to \(X = 12\).

\[
P(X \leq 12) = \sum_{k=0}^{12} P(X = k)
\]

#### 2. The probability that more than 14 subscribers are not satisfied
We need to calculate \(P(X > 14)\), which is the complement of \(P(X \leq 14)\):
\[
P(X > 14) = 1 - P(X \leq 14)
\]

#### 3. The probability that between 14 and 18 subscribers are not satisfied
We need to calculate \(P(14 \leq X \leq 18)\), which is:
\[
P(14 \leq X \leq 18) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)
\]

#### 4. The probability that exactly 18 subscribers are not satisfied
This is just \(P(X = 18)\).

Let me calculate all of these probabilities for you.Here are the results of the calculations:

### (a) For a sample of 6 subscribers:
The probability that 4 or more subscribers are not satisfied with their service is:
\[
P(X \geq 4) = 0.9011
\]

### (b) For a sample of 20 subscribers:
1. The probability that 12 or fewer subscribers are not satisfied:
\[
P(X \leq 12) = 0.0321
\]

2. The probability that more than 14 subscribers are not satisfied:
\[
P(X > 14) = 0.8042
\]

3. The probability that between 14 and 18 subscribers are not satisfied:
\[
P(14 \leq X \leq 18) = 0.8441
\]

4. The probability that exactly 18 subscribers are not satisfied:
\[
P(X = 18) = 0.1369
\]

If you'd like further details on these probabilities, feel free to ask!

Here are five related questions:
1. How does the binomial distribution change with larger sample sizes?
2. What is the effect of decreasing the probability \(p\) on the distribution?
3. How would the probabilities change if the sample size were increased to 30?
4. Can we use a normal approximation for large sample sizes in this case?
5. How would these probabilities differ if we asked for "satisfaction" instead of "not satisfied"?

**Tip:** For larger sample sizes, the binomial distribution can be approximated by the normal distribution when conditions are met.

A consumer advocate claims that 80% 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 6 cable subscribers is selected. Calculate the probability that 4 or more subscribers in the sample are not satisfied with their service. 

(b) Suppose that the advocate's claim is true, and suppose that a random sample of 20 cable subscribers is selected.
1. The probability that 12 or fewer subscribers in the sample are not satisfied with their service.
2. The probability that more than 14 subscribers in the sample are not satisfied with their service.
3. The probability that between 14 and 18 (inclusive) subscribers in the sample are not satisfied with their service.
4. The probability that exactly 18 subscribers are not satisfied with their service.

Calculate probabilities using the binomial distribution for cable television subscribers' satisfaction rates. Explore the advocate's claim that 80% of subscribers are not satisfied, with scenarios for sample sizes of 6 and 20. This problem includes detailed probability calculations and is suitable for advanced high school or college students.

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