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More on insurance An insurance company knows that in the entire population of homeowners, the mean annual loss from 铿乺e is =250and the standard deviation of the loss is =300. The distribution of losses is strongly right-skewed: many policies have 0loss, but a few have large losses. If the company sells 10,000 policies, can it safely base its rates on the assumption that its average loss will be no greater than 275? Follow the four-step process

Short Answer

Expert verified

From the given information, the company can safely base its rates on the assumption that is average loss will be no longer greater than275 it sells,10,000 policies.

Step by step solution

01

Given Information

It is given in the question that, the mean annual loss, =250

the standard deviation of the loss, =300

follow the four-step process.

02

Explanation

The central limit theorem states that if the sample size of a sampling distribution is 300or more, then the sample mean is approximately normal whose mean is and the standard deviation is n.

The zvalue of a distribution can be found by dividing the difference between the population mean and sample mean by the standard deviation that is, z=xsrn.

Since the sample size of 10,000policies is at least 30; so we can apply the central limit theorem.

Find the zvalue by using the formulaz=xsrn.

03

Explanation

Substitute 275for x,250for,300for,and10,000fornthe above formula and simplify.

z=27525030010000

=25306100

=8.33

Thus, the corresponding probability is:

P(x>275)=P(z>8.33)=P(Z<8.33)=0.0001

Thus, the company can safely assume that the average loss will be no greater than 275because the probability is almost zero.

Accordingly, the company can safely base its rates on the assumption that its average loss will be no greater than 275it sells 10,000policies.

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Most popular questions from this chapter

Doing homework A school newspaper article claims that 60%of the students at a large high school did all their assigned homework last week. Some skeptical AP Statistics students want to investigate whether this claim is true, so they choose an SRS of 100students from the school to interview. What values of the sample proportion p藛 would be consistent with the claim that the population proportion of students who completed all their homework is p 0.60? To find out, we used Fathom software to simulate choosing 250SRSs of size n=100students from a population in which p=0.60. The figure below is a dotplot of the sample proportion pof students who did all their homework.

(a) Is this the sampling distribution of p? Justify your answer.

(b) Describe the distribution. Are there any obvious outliers?

(c) Suppose that 45of the100students in the actual sample say that they did all their homework last week. What would you conclude about the newspaper article鈥檚 claim? Explain.

Suppose that you are a student aide in the library and agree to be paid according to the 鈥渞andom pay鈥 system. Each week, the librarian flips a coin. If the coin comes up heads, your pay for the week is \(80. If it comes up tails, your pay for the week is \)40. You work for the library for 100 weeks. Suppose we choose an SRS of 2 weeks and calculate your average earnings x. The shape of the sampling distribution of xwill be

(a) Normal.

(b) approximately Normal

(c) right-skewed

(d) left-skewed.

(e) symmetric but not Normal.

Do you go to church? What sample size would be required to reduce the standard deviation of the sampling distribution to one-third the value you found in Exercise 36 (b)? Justify your answer.

Do you drink cereal milk? What sample size would be required to reduce the standard deviation of the sampling distribution to one-half the value you found in Exercise 35(b)? Justify your answer.

Do you jog? The Gallup Poll once asked a random sample of1540adults, "Do you happen to jog?" Suppose that in fact 15%of all adults jog.

(a) What is the mean of the sampling distribution of p^? Justify your answer.

(b) Find the standard deviation of the sampling distribution of p^. Check that the 10%condition is met.

(c) Is the sampling distribution of p^approximately Normal? Justify your answer.

(d) Find the probability that between 13%and 17%of a random sample of 1540adults are joggers. Show your work.

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