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Use Table A to 铿乶d the proportion of observations from the standard Normal distribution that satis铿乪s each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question. Use your calculator or the Normal Curve applet to check your answers.

More Table A practice

(a) zis between 鈭2.05and 0.78

(b) zis between 鈭1.11and 鈭0.32

Short Answer

Expert verified

From the given information

a) The area between z=-2.05&z=0.78is 0.7823-0.0202=0.7621.

b) The area between z=-1.11&z=-0.32is0.3745-0.1335=0.2410

Step by step solution

01

Part (a) Step 1: Given Information

It is given in the question that,

zis between -2.05and 0.78

02

Part (a) Step 2: Explanation

The below Standard Normal probabilities table is a table of areas under the standard Normal Curve. The table entry for each value zis the area under the curve to the left of z.

03

Part (a) Step 3: Graphical Representation

Shows the graph and table.

04

Part (a) Step 4: Explanation

(a) A standard normal probability table shows the area to the left ofz=0.78is 0.7823and the area to the left of z=-2.05is 0.0202.

Between z=-2.05and z=0.78there is a mass that is half of what is to the left of0.78minus the area to the left of -2.05.

Area between z=-2.05&z=0.78 is:

0.7823-0.0202=0.7621

By the graphical representation , the result will be,

05

Part (b) Step 1: Given Information

It is given in the question that,

z is between -1.11and -0.32

06

Part (b) Step 2: Explanation

(b) A standard normal probability table shows that the area to the left of z=-0.32is 0.3745and the area to the left of z=-1.11is 0.1335.

Between z=-1.11and z=-0.32there is a mass that is half of what is to the left of -0.32minus the area to the left of -1.11.

Area between z=-1.11&z=-0.32:

0.3745-0.1335=0.2410

By the graphical representation , the result will be,

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Most popular questions from this chapter

The weights of laboratory cockroaches follow a Normal distribution with mean 80grams and standard deviation 2grams. The figure below is the Normal curve for this distribution of weights.

Point C on this Normal curve corresponds to

(a) 84grams.

(c) 78grams.

(e) 74grams

(b) 82grams.

(d) 76grams.

Follow the method shown in the examples to answer each of the following questions. Use your calculator or the Normal Curve applet to check your answers.

2. People with cholesterol levels between 200and 240mg/dlare at considerable risk for heart disease. What percent of 14-year-old boys have blood cholesterol between 200and 240mg/dl?

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 was 10.9Nationally, 5.2% of test-takers earned a score of 65 or higher on the Critical Reading test鈥檚 20 to 80 scale. 9

PSAT scores How well did the boys at Scott鈥檚 school perform on the PSAT? Give appropriate evidence to support your answer.

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.

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 515 and the standard deviation was 116 .
T2.9. What is the standardized score ( z-score) for a student who scored 500 on the old SAT scale?
(a) -30
(b) -0.27
(c) -0.13
(d) 0.13
(e) 0.27

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