{Entrance exam narrative} [Normal table or calculator required.] Bob's original score was 130 and Jill's standard score was +1.5.What percentage of the students taking this exam had scores that fell between Bob and Sue's scores?
A)16%
B)93%
C)77%
D)Not enough information to tell.

Question

Quizplus · Accepted Answer

Without knowing Sue's score, we can still find the percentage of students who scored between Bob and Sue's score by using the normal distribution table or calculator. Assuming a normal distribution, we can convert Bob's original score of 130 to a standard score using the formula z = (x - μ) / σ. Given that the mean (μ) of the scores is not provided, we can assume a standard mean of 100 and a standard deviation (σ) of 15 (typical for many standardized tests). Thus, Bob's standard score is (130 - 100) / 15 = 2.

Next, we can use the Z-table or calculator to find the percentage of test-takers who scored between a standard score of 2 and 1.5. The area to the left of a standard score of 2 is 0.9772 and the area to the left of a standard score of 1.5 is 0.9332. Therefore, the percentage of students who scored between Bob and Jill's scores is 0.9772 - 0.9332 = 0.044 or 4.4%.

To find the percentage of students who scored between Bob and Sue's scores, we would need to know Sue's score to calculate her standard score and then find the difference between her score and Bob's score in terms of standard deviation units.

{Entrance Exam Narrative} [Normal Table or Calculator Required