MGSC 1108 Solutions Final Exam Outline Spring.docx - MGSC 5108 Problem Solutions Chapter 6 23 a b c 24 a b c 28 0.1525 found by subtracting 0.4938 \u2013

# MGSC 1108 Solutions Final Exam Outline Spring.docx - MGSC...

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MGSC 5108 Problem Solutions Chapter 6 23. a. 0.1525, found by subtracting 0.4938 – 0.3413, which are the areas associated with z values of 2.5 and 1, respectively. b. 0.0062, found by 0.5000 – 0.4938 c. 0.9710, found by recalling that the area of the z value of 2.5 is 0.4938. Then find z = – 2.00 found by ((205 – 225)/10). Thus, 0.4938 + 0.4772 = 0.9710. 24. a. 0.3085, found by z = (\$80,000 – \$70,000)/\$20,000 = 0.50. The area is 0.1915. Then 0.5000 – 0.1915 = 0.3085 b. 0.2902, found by z = ((80,000 – 70,000)/20,000) = 0.50, the area is 0.1915 z = ((65,000 – 70,000)/20,000) = – 0.25, the area is 0.0987 Adding these values together: 0.1915 + 0.0987 = 0.2902 c. 0.5987, found by the area under the curve with a z = – 0.25, 0.0987 + 0.5000 = 0.5987 You must first find the z-score corresponding to .80/2 =.40 of the normal curve. Z = -.84 28. 084 80 14 80 00 1176 68 24 . . . . X X 29. 200.7; find a z value where 0.4900 of area is between 0 and z . That value is z = 2.33, then solve for X : ( X – 200)/0.3 so X = 200.7 31. 1630, found by 2100 1.88(250) (First find the z-score corresponding to .47 of the curve. From the table, this is z=1.88. Then find x.) Chapter 7 16. a. z = 74 75 1.26 5/ 40  So probability is 0.1038, found by 0.5000 0.3962. b. z = 76 75 1.26 5/ 40 So probability is 0.7924, found by 2(0.3962).

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