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The mean of this distribution (don’t try to find it) will be

a. very close to the median.

b. greater than the median.

c. less than the median.

d. You can’t say, because the distribution isn’t symmetric.

e. You can’t say, because the distribution isn’t Normal.

Short Answer

Expert verified

Option C: The mean of this distribution (don’t try to find it) will be less than the median.

Step by step solution

01

Given information

The following options are

a. very close to the median.

b. greater than the median.

c. less than the median.

d. You can’t say, because the distribution isn’t symmetric.

e. You can’t say, because the distribution isn’t Normal.

02

Explanation for correct option

Because most dots are on the right side of the dot and there is a tail of fewer dots to the left in the dot plot, the dot plot is skewed to the left. The uncommon values have little impact on the median. The mean would be smaller than the median since there are unusually small values.

As a result, the best solution is (c)

03

Explanation for incorrect option

(a)The mean of this distribution (don’t try to find it) will not be very close to the median.

(b) The mean of this distribution (don’t try to find it) will not be greater than the median.

(d)The mean of this distribution (don’t try to find it) will not be you can’t say, because the distribution isn’t symmetric.

(e) The mean of this distribution (don’t try to find it) will not be you can’t say, because the distribution isn’t Normal.

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Most popular questions from this chapter

An insurance company claims that in the entire population of homeowners, the mean annual loss from fire is and the standard deviation of the loss is σ=\(5000.The distribution of losses is strongly right-skewed: many policies have \)0loss, but a few have large losses. The company hopes to sell 1000 of these policies for \(300each.

a. Assuming that the company’s claim is true, what is the probability that the mean loss from fire is greater than \)300for an SRS of 1000 homeowners?

b. If the company wants to be 90% certain that the mean loss from fire in an SRS of 1000 homeowners is less than the amount it charges for the policy, how much should the company charge?

A newborn baby has extremely low birth weight (ELBW) if it weighs less than 1000 grams. A study of the health of such children in later years examined a random sample of 219 children. Their mean weight at birth was x-x¯=810grams. This sample mean is an unbiased estimator of the mean weight μ in the population of all ELBW babies, which means that

a. in all possible samples of size 219 from this population, the mean of the values of x-x¯will equal 810 .

b. in all possible samples of size 219 from this population, the mean of the values of x-x¯will equal μ

c. as we take larger and larger samples from this population, x-x¯will get closer and closer to μ.

d. in all possible samples of size 219 from this population, the values of x-x¯will have a distribution that is close to Normal.

e. the person measuring the children's weights does so without any error.

Dem bones (2.2) Osteoporosis is a condition in which the bones become brittle due to the loss of minerals. To diagnose osteoporosis, an elaborate apparatus measures bone mineral density (BMD). BMD is usually reported in a standardized form. The standardization is based on a population of healthy young adults. The World Health Organization (WHO) criterion for osteoporosis is a BMD score that is 2.5standard deviations below the mean for young adults. BMD measurements in a population of people similar in age and gender roughly follow a Normal distribution.

a. What percent of healthy young adults have osteoporosis by the WHO criterion?

b. Women aged 70to 79are, of course, not young adults. The mean BMD in this age group is about-2 on the standard scale for young adults. Suppose that the standard deviation is the same as for young adults. What percent of this older population has osteoporosis?

10%Why is it important to check the 10 % condition before calculating probabilities involving localid="1654670795605">x-?

a. To reduce the variability of the sampling distribution of x-x¯

b. To ensure that the distribution of x-x¯is approximately Normal

c. To ensure that we can generalize the results to a larger population

d. To ensure that x-x¯will be an unbiased estimator of μ

e. To ensure that the observations in the sample are close to independent

More homework Some skeptical Ap® Statistics students want to investigate the newspaper's claim in Exercise 11, so they choose an SRS of 100students from the school to interview. In their sample, 45students completed their homework last week. Does this provide convincing evidence that less than 60%of all students at the school completed their assigned homework last week?

a. What is the evidence that less than 60%of all students completed their assigned homework last week?

b. Provide two explanations for the evidence described in part (a).

We used technology to simulate choosing 250SRSs of size n=100n=100from a population of 2000students where 60%completed their assigned homework last week. The dotplot shows pp^the sample proportion of students who completed their assigned homework last week for each of the 250simulated samples.

c. There is one dot on the graph at 0.73. Explain what this value represents.

d. Would it be surprising to get a sample proportion of p=0.45p^=0.45or smaller in an SRS of size 100when p=0.60p=0.60? Justify your answer.

e. Based on your previous answers, is there convincing evidence that less than 60%of all students at the school completed their assigned homework last week? Explain your reasoning.

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