(4.2.20) An insurance company offers a discount to homeowners who install smoke detectors in their homes. A company representative claims that 80% or more of policyholders have smoke detectors. You draw a random sample of eight policyholders. Let X be the number of policyholders in the sample who have smoke detectors. a. If exactly 80% of the policyholders have smoke detectors, what is P(X ≤ 1)? b. Based on the answer to part (a), if 80% of the policyholders have smoke detectors, would one policyholder with a smoke detector in a sample of size 8 be an unusual small number? Explain. c. If you found that one of the eight sample policyholders had a smoke detector, would this be convincing evidence that the claim is false? Explain. d. If exactly 80% of the policyholders have smoke detectors, what is P(X ≤ 6)? e. Based on the answer to part (d), if 80% of the policyholders have smoke detectors, would six policy holders with smoke detectors in a sample of size 8 be an unusually small number? Explain. f. If you found that six of the eight sample policyholders had smoke detectors, would this be convincing evidence that the claim is false? Explain.

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Step-by-step explanation:

The proportion that have smoke detectors = p = 80% = 0.80

In a sample of 8 policyholders,

Mean = np = 8×0.8 = 6.4

Standard deviation = √[np(1-p)] = √[8×0.8×0.2] = 1.13

a) P(X ≤ 1)

We first normalize 1.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (1 - 6.4)/1.13 = - 4.78

To determine the probability of less than or equal to 1 policyholder with smoke detectors.

P(X ≤ 1) = P(z ≤ -4.78)

We'll use data from the normal probability table for these probabilities

P(X ≤ 1) = P(z ≤ -4.78) = 0.000

b) Based on the answer to part (a), if 80% of the policyholders have smoke detectors, would one policyholder with a smoke detector in a sample of size 8 be an unusual small number? Explain.

Based on the probability obtained P(X ≤ 1) = 0.00, it is evident that it is very unlikely that in a sample of 8 policyholders, there is only 1 policyholder with a smoke detector.

It would be very unusual because according to the mathematical probability, it shouldn't happen at all.

c) If only one policyholder has smoke detectors in a sample of 8 policyholders, it would seriously throw the claim that 80% of the population of policyholders have smoke detectors.

The largest confidence interval for where the means for any sample can be obtained using 99.99% would explain this the best.

Confidence interval = (Sample mean) ± (Margin of error)

Sample mean = 6.4

margin of error = (critical value) × (standard deviation)

Critical value = 3.891 for 99.99% confidence level.

Standard deviation = 1.13

margin of error = 3.891 × 1.13 = 4.4

Confidence interval = (Sample mean) ± (Margin of error)

Confidence interval = 6.4 ± 4.4 = (2.00, 10.80) = (2, 8)

A sample of 8 yielding only 1 policyholder with smoke detector is outside one of the largest confidence intervals, hence, it shows more how unusual and how much this would throw the origin so claim into doubt.

d) P(X ≤ 6)

We first normalize 6

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (6 - 6.4)/1.13 = - 0.35

To determine the probability of less than or equal to 6 policyholder with smoke detectors.

P(X ≤ 6) = P(z ≤ -0.35)

We'll use data from the normal probability table for these probabilities

P(X ≤ 6) = P(z ≤ -0.35) = 0.36317

e) Based on the answer to part (d), if 80% of the policyholders have smoke detectors, would six policy holders with smoke detectors in a sample of size 8 be an unusually small number? Explain.

Based on the answer for the probability obtained, six policy holders with smoke detectors in a sample of size 8 would not represent an unusually small number. A probability of 0.36317 is considerable enough.

f) Six of the eight sample policyholders having smoke detectors, would this be convincing evidence that the claim is false? Explain.

No it wouldn't. The number 6, sits properly in almost all of the confidence intervals that can be computed for this sample size. So, it would in no way put the claim in doubt.

Hope this Helps!!!

i'm sorry i don't know

step-by-step explanation:

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