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Chapter 2: Q 3.3. (page 98)

Now suppose that you convert the class’s heights to z-scores. What would be the shape, center, and spread of this distribution? Explain.

Short Answer

Expert verified

Mean will become 0 (zero) and standard deviation would be 1 (one)

Step by step solution

01

Step 1. Given

02

Step 2. Concept

The following Minitab computer output validates the outcome: When each observation in distribution is standardized, the resultant set of z-scores has a mean of 0 and a standard deviation of 1

03

Step 3. Explanation

Height is now measured in z-scores. The shape will not be modified by the transformation from height to z-scores, therefore the shape will not be changed, but the mean will be shifted to 0 (zero) and the standard deviation will become1based on the given data (one)

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Most popular questions from this chapter

Each year, about 1.5million college-bound high school juniors take the PSAT. In a recent year, the mean score on the Critical Reading test was 46.9and the standard deviation was10.9. Nationally, 5.2%of test takers earned a score of 65 or higher on the Critical Reading test’s 20to 80scale.9

PSAT scores Scott was one of 50junior boys to take the PSAT at his school. He scored 64on the Critical Reading test. This placed Scott at the 68th percentile within the group of boys. Looking at all 50boys’ Critical Reading scores, the mean was 58.2and the standard deviation was 9.4

(a) Write a sentence or two comparing Scott’s percentile among the national group of test-takers and among the 50boys at his school.

(b) Calculate and compare Scott’s z-score among these same two groups of test-takers.

George has an average bowling score of 180 and bowls in a league where the average for all bowlers is 150 and the standard deviation is 20 Bill has an average bowling score of 190 and bowls in a league where the average is 160 and the standard deviation is 15 Who ranks higher in his own league, George or Bill?

(a) Bill, because his 190 is higher than George’s 180

(b) Bill, because his standardized score is higher than George’s.

(c) Bill and George have the same rank in their leagues because both are 30 pins above the mean.

(d) George, because his standardized score is higher than Bill’s.

(e) George, because the standard deviation of bowling scores is higher in his league.

Use Table A in the back of the book to find the proportion of observations from a standard Normal distribution that fall in each of the following regions. In each case, sketch a standard Normal curve and shade the area representing the region.

2.z>−2.15

What the mean means The figure below is a density curve. Trace the curve onto your paper.

(a) Mark the approximate location of the median. Justify your choice of location.

(b) Mark the approximate location of the mean. Justify your choice of location.

Use the 68-95-99.7 rule to estimate the percent of observations from a Normal distribution that fall in an interval involving points one, two, or three standard deviations on either side of the mean.

Questions T2.9 and T2.10 refer to the following setting. Until the scale was changed in 1995, SAT scores were based on a scale set many years ago. For Math scores, the mean under the old scale in the 1990swas 470and the standard deviation was 110. In 2009, the mean was 515and the standard deviation was 116 .
T2.10. Jane took the SAT in 1994and scored 500. Her sister Colleen took the SAT in 2009and scored 530. Who did better on the exam, and how can you tell?
(a) Colleen-she scored 30 points higher than Jane.
(b) Colleen-her standardized score is higher than Jane's.
(c) Jane-her standardized score is higher than Colleen's.
(d) Jane-the standard deviation was bigger in 2009.
(e) The two sisters did equally well-their z-scores are the same.
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