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...

This preview shows page 1 - 2 out of 5 pages.

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).
Image of page 1
Image of page 2

You've reached the end of your free preview.

Want to read all 5 pages?

  • Winter '18

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes
A+ icon
Ask Expert Tutors