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Chapter 6: Q.26 (page 356)

Friends How many close friends do you have? An opinion poll asks this question of an SRS of 1100 adults. Suppose that the number of close friends adults claim to have varies from person to person with mean μ=9 and standard deviation σ=2.5. We will see later that in repeated random samples of size 1100, the mean response x will vary according to the Normal distribution with mean 9 and standard deviation 0.075. What is role="math" localid="1649504967961" P(8.9≤x≤9.1), the probability that the sample result x estimates the population truth μ=9 to within ±0.1?

Short Answer

Expert verified

The probability ofP(8.9≤x≤9.1)is0.8164.

Step by step solution

01

Given Information

An opinion poll asks this question of an SRS of 1100 adults. Suppose that the number of close friends adults claim to have varies from person to person with mean μ=9 and standard deviation σ=2.5.

02

Given Information

The sample mean is normally distributed with mean μand standard deviation σ/n.

Thez-score is the sample value multiplied by the standard deviation divided by the mean:

localid="1649915865099" z=x¯-μσ/n=8.9-92.5/1100≈-1.33

localid="1649915875646" z=x¯-μσ/n=9.1-92.5/1100≈1.33

Using the normal probability table in the appendix, calculate the corresponding probability:

localid="1649916024235" P(8.9≤x¯≤9.1)=P(-1.33<Z<1.33)=P(Z<1.33)-P(Z<-1.33)=0.9082−0.0918=0.8164

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Most popular questions from this chapter

A large auto dealership keeps track of sales and leases agreements made during each hour of the day. Let χ= the number of cars sold and γ= the number of cars leased during the first hour of business on a randomly selected Friday. Based on previous records, the probability distributions of χand γare as follows:

Define τ=χ+γ

Compute στassuming that χ and γ are independent. Show your work.

For each of the following situations, determine whether the given random variable has a binomial distribution. Justify your answer.

Choose three students at random from your class. Let Y=the number who are over 6feet tall.

Life insurance A life insurance company sells a term insurance policy to a21-year-old male that pays 100,000if the insured dies within the next 5years. The probability that a randomly chosen male will die each year can be found in mortality tables. The company collects a premium of 250each year as payment for the insurance. The amount Y that the company earns on this policy is 250per year, less the 100,000that it must pay if the insured dies. Here is a partially completed table that shows information about risk of mortality and the values of Y=profit earned by the company:

(a) Copy the table onto your paper. Fill in the missing values of Y.

(b) Find the missing probability. Show your work.

(c) Calculate the mean μY. Interpret this value in context.

18. Life insurance

(a) It would be quite risky for you to insure the life of a 21-year-old friend under the terms of Exercise 14. There is a high probability that your friend would live and you would gain \(1250in premiums. But if he were to die, you would lose almost \)100,000. Explain carefully why selling insurance is not risky for an insurance company that insures many thousands of 21-year-old men.

(b) The risk of an investment is often measured by the standard deviation of the return on the investment. The more variable the return is, the riskier the
investment. We can measure the great risk of insuring a single person’s life in Exercise 14by computing the standard deviation of the income Y that the insurer will receive. Find σY using the distribution and mean found in Exercise 14.

A large auto dealership keeps track of sales and leases agreements made during each hour of the day. Let χ= the number of cars sold and γ= the number of cars leased during the first hour of business on a randomly selected Friday. Based on previous records, the probability distributions of χand γare as follows:

DefineD=X-Y

Find and interpret μD.

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