Question

Mr. Pryor's second period consists of 25 students. The mean grade on the first exam was 80, the median was also 80, and the standard deviation was 6.07. Find the standardized scores (z-scores) for each of the following students. Interpret each value in context.

a) Katie, who scored 93.
b) Norman, who scored 72
c) Ted, who scored an 80.
d) Jenny earned an 82 on Mr. Goldstone's chemistry test. Mr. Goldstone told the class that the distribution was fairly symmetric with a mean of 76 and a standard deviation of 4. She scored an 86 in Mr. Pryor's exam, on which test did Jenny perform better relative to the class? Justify your answer.

Answer 1

See explanation

Explanation:

a. (93-80)/6.07=

13/6.07=2.14

b. (72-80)/6.07=

8/6.07=1.317957

1.317957=1.32

c. (80-80)/6.07=

0/6.07=0

d. (82-76)/4=

6/4=1.5

(86-80)/6.07=

6/6.07=0.988

1.5>0.98, so Jenny did better on Mr. Goldstone's test.