QUIZ_2 solutions

QUIZ_2 solutions - QUIZ 2 Solutions#3.13 Page 83 MATH 1070...

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Unformatted text preview: QUIZ 2 Solutions #3.13. Page 83. MATH 1070 Spring 08 Use Table A to find the Z of a standard Normal variable that satisfies each of the following conditions. (Use the value of Z to from Table A [Page 684] that comes closest to satisfying the condition.) In each case, sketch a Standard Normal Curve with your value of Z marked on the axis. a. The point Z with 25% of the observations falling below it. b. The point Z with 40% of the observations falling above it. a.) Solution: This Z result can be found in the ZTable by retrieving the Z associated with the 25% =0.25 proportion. The Z associated with this is proportion is Z=0.6745. To use the TI83/84, use the following steps: 2nd VAR=>invNorm(0.25)=0.6745 b.) Solution This Z score can be found in the ZTable by retrieving the Z associated with the 60% =0.60 proportion. This is also equivalent to finding the Z score associated with the 40% proportion to the right of this Z score. The Z score associated with this proportion is Z=0.2533. To use the TI83/84, use the following steps: 2nd VAR=>invNorm(0.60)=0.2533 #3.14. Page 83 Scores on the Wechsler Adult Intelligence Scale are approximately Normally distributed with =100 and =15. a. What scores fall in the lowest 25% of the distribution? b. How high a score is needed to be in the highest 5% a.) Solution: This Z result can be found in the ZTable by retrieving the Z associated with the 25% =0.25 proportion. The Z associated with this is proportion is Z=0.6745. To use the TI83/84, use the following steps: 2nd VAR=>invNorm(0.25)=0.6745 Now using the formula X = Z + X=0.6745*15+100 X=89.88. This implies that X 89.88 for the score to lie in the lowest 25% of the distribution. b.) Solution This Z result can be found in the ZTable by retrieving the Z associated with the 95% =0.95 proportion. . This is equivalent to finding the Z score associated with the top 5% proportion i.e to the right of this Z score. The Z score associated with this proportion is Z=1.65. To use the TI83/84, following the following steps: 2nd VAR=>invNorm(0.95)=1.65 Now using the formula X = Z + X=1.65*15+100 X=124.75. This implies that if one makes a score that is X the exam takers. 124.75 then he/she is in the highest 5% of ...
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This note was uploaded on 04/30/2008 for the course MATH 1070 taught by Professor Akbas during the Spring '08 term at Georgia State.

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