Stats: Modeling the World Nasta Edition Grades 9-12
Stats: Modeling the World Nasta Edition Grades 9-12
3rd Edition
ISBN: 9780131359581
Author: David E. Bock, Paul F. Velleman, Richard D. De Veaux
Publisher: PEARSON
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Chapter 6, Problem 15E

a)

To determine

To find which student qualifies.

a)

Expert Solution
Check Mark

Answer to Problem 15E

Megan qualifies for language honors.

Explanation of Solution

Given:

French exam:

Mean = μ = 81

Standard deviation = σ =5

Spanish exam:

Mean = μ = 74

Standard deviation = σ =15

Anna’s scores:

  x1 =83

  x2 = 83

Megan’s scores:

  x1 = 77

  x2 = 95

Average score of Anna:

  83+832=83

Average score of Megan:

  95+772=86

The average score of Megan is high, therefore Megan qualifies for language honors.

b)

To determine

To explain which student’s performance was better.

b)

Expert Solution
Check Mark

Answer to Problem 15E

Anna did better performance.

Explanation of Solution

Given:

French exam:

Mean = μ = 81

Standard deviation = σ =5

Spanish exam:

Mean = μ = 74

Standard deviation = σ =15

Anna’s scores:

  x1 =83

  x2 = 83

Megan’s scores:

  x1 = 77

  x2 = 95

Formula:

  z=xμσ

We need to take Sum of z-scores to judge a performance.

Sum z-score of Anna:

  z=83815+837415=1

Sum z-score of Megan:

  z=77815+957415=0.6

Average z-score of Anna is higher than Megan. Hence, Anna did better performance.

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