Mathematics College

Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. AIB Insurance randomly sampled 100 recently paid policies and determined the average age of clients in this sample to be 77.7 years with a standard deviation of 3.6. The 90% confidence interval for the true mean age of its life insurance policy holders is
A. (76.87, 80.33)
B. (72.5, 82.9)
C. (77.1, 78.3)
D. (74.1, 81.3)
E. (74.5, 80)

Answers

Answer 1

Answer:

[tex]77.7-1.66\frac{3.6}{\sqrt{100}} =77.102[/tex]    

[tex]77.7+1.66\frac{3.6}{\sqrt{100}} =78.30[/tex]    

And the best option would be:

C. (77.1, 78.3)

Step-by-step explanation:

Information given

[tex]\bar X=77.7[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=3.6 represent the sample standard deviation

n=100 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

The degrees of freedom are given by:

[tex]df=n-1=100-1=99[/tex]

Since the Confidence is 0.90 or 90%, the significance would be [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and the critical value for this case would be [tex]t_{\alpha/2}=1.66[/tex]

And replacing we got:

[tex]77.7-1.66\frac{3.6}{\sqrt{100}} =77.10[/tex]    

[tex]77.7+1.66\frac{3.6}{\sqrt{100}} =78.30[/tex]    

And the best option would be:

C. (77.1, 78.3)

Related Questions

In a certain community, eight percent of all adults over age 50 have diabetes. If a health service in this community correctly diagnosis 95% of all persons with diabetes as having the disease and incorrectly diagnoses ten percent of all persons without diabetes as having the disease, find the probabilities that:

Answers

Complete question is;

In a certain community, 8% of all people above 50 years of age have diabetes. A health service in this community correctly diagnoses 95% of all person with diabetes as having the disease, and incorrectly diagnoses 10% of all person without diabetes as having the disease. Find the probability that a person randomly selected from among all people of age above 50 and diagnosed by the health service as having diabetes actually has the disease.

Answer:

P(has diabetes | positive) = 0.442

Step-by-step explanation:

Probability of having diabetes and being positive is;

P(positive & has diabetes) = P(has diabetes) × P(positive | has diabetes)

We are told 8% or 0.08 have diabetes and there's a correct diagnosis of 95% of all the persons with diabetes having the disease.

Thus;

P(positive & has diabetes) = 0.08 × 0.95 = 0.076

P(negative & has diabetes) = P(has diabetes) × (1 –P(positive | has diabetes)) = 0.08 × (1 - 0.95)

P(negative & has diabetes) = 0.004

P(positive & no diabetes) = P(no diabetes) × P(positive | no diabetes)

We are told that there is an incorrect diagnoses of 10% of all persons without diabetes as having the disease

Thus;

P(positive & no diabetes) = 0.92 × 0.1 = 0.092

P(negative &no diabetes) =P(no diabetes) × (1 –P(positive | no diabetes)) = 0.92 × (1 - 0.1)

P(negative &no diabetes) = 0.828

Probability that a person selected having diabetes actually has the disease is;

P(has diabetes | positive) =P(positive & has diabetes) / P(positive)

P(positive) = 0.08 + P(positive & no diabetes)

P(positive) = 0.08 + 0.092 = 0.172

P(has diabetes | positive) = 0.076/0.172 = 0.442

The probability are "0.168 and 0.452".

Using formula:

[tex]P(\text{diabetes diagnosis})\\[/tex]:

[tex]=\text{P(having diabetes and have been diagnosed with it)}\\ + \text{P(not have diabetes and yet be diagnosed with diabetes)}[/tex]

[tex]=0.08 \times 0.95+(1-0.08) \times 0.10 \\\\=0.08 \times 0.95+0.92 \times 0.10 \\\\=0.076+0.092\\\\=0.168[/tex]

[tex]\text{P(have been diagnosed with diabetes)}[/tex]:

[tex]=\frac{\text{P(have diabetic and been diagnosed as having insulin)}}{\text{P(diabetes diagnosis)}}[/tex]

[tex]=\frac{0.08\times 0.95}{0.168} \\\\=\frac{0.076}{0.168} \\\\=0.452\\[/tex]

Learn more about the probability:

brainly.com/question/18849788

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