Question

1. Ted who took the final exam scored 87 on the exam A and 92 on the exam B. The mean score for exam A for all test takers was 75 with a standard deviation of 5, and the mean score for the exam B was 80 with a standard deviation of 6. Suppose that both distributions are nearly normal.

(a) Write down the short-hand for these two normal distributions.

(b) What is Ted's Z-score on the exam A and B? Draw a standard normal distribution curve and mark these two Z-scores.

(c) What do these Z-scores tell you?

(d) Relative to others, which section did she do better on?

(e) Find her percentile scores for the two exams.

(f) What percent of the test takers did better than her on the exam A? On the exam B?

(g) Explain why simply comparing raw scores from the two sections could lead to an incorrect conclusion as to which section a student did better on.

Answer 1

a) X_A ~ N( 75, 5² ) For exam A

X_B ~ N( 80, 6²) For the exam B

b) Z score in exam A,

Z score in exam B,

c) The z score for exam A says that Ted is 2.4 standard deviation far from the mean score in exam A

And the z score for Exam B says that ted is 2 standard deviation far ahead from the mean score of the exam

d) Since Z_A > Z_B, Ted do well in exam A

e) Percentile score for exam A, P( Z < 2.4) = 0.9918

Percentile score for exam B , P( Z < 2) = 0.97725

f) For Exam A : (1-0.9918)= 0.0082 = 0.82% of the students did better than her on exam A

  For Exam B: (1-0.97725)= 0.02275 = 2.275% of the students did better than her on exam B

g) Simply comparing raw scores from the two sections could lead to an incorrect conclusion as to which section a student did better on because this not only depend on their score bur relative score of the other students in the exam. Thats why we use z score to compare.