Chapter 6, Problem 15E

a)

To determine

To find which student qualifies.

Answer to Problem 15E

Megan qualifies for language honors.

Explanation of Solution

Given:

French exam:

Mean = $\mu$ = 81

Standard deviation = $\sigma$ =5

Spanish exam:

Mean = $\mu$ = 74

Standard deviation = $\sigma$ =15

Anna’s scores:

  $x_1$ =83

  $x_2$ = 83

Megan’s scores:

  $x_1$ = 77

  $x_2$ = 95

Average score of Anna:

  $\frac{83+83}{2}=83$

Average score of Megan:

  $\frac{95+77}{2}=86$

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

b)

To determine

To explain which student’s performance was better.

Answer to Problem 15E

Anna did better performance.

Explanation of Solution

Given:

French exam:

Mean = $\mu$ = 81

Standard deviation = $\sigma$ =5

Spanish exam:

Mean = $\mu$ = 74

Standard deviation = $\sigma$ =15

Anna’s scores:

  $x_1$ =83

  $x_2$ = 83

Megan’s scores:

  $x_1$ = 77

  $x_2$ = 95

Formula:

  $z=\frac{x-\mu }{\sigma }$

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

Sum z-score of Anna:

  $z=\frac{83-81}{5}+\frac{83-74}{15}=1$

Sum z-score of Megan:

  $z=\frac{77-81}{5}+\frac{95-74}{15}=0.6$

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