Question

Important: please show work for each question. Thank you!

In a previous section of PSY230, the second exam was worth 80 points. The scores from that class were normally distributed with a mean (μ ) of 65 and a standard deviation (σ) of 5. If the exam scores were converted to a Z distribution, the distribution would form a perfect bell shape. The following questions require locating individual exam scores on the Z distribution and examine the percentage (or proportion) of cases above or below a score.

Hints: It helps to draw a Z distribution (bell curve) and place John’s and Tom’s Z scores on the distribution for answering the questions. Use the Z table for converting between Z score and area (percentage) of the distribution.

1. John obtained a score of 74. What is John’s z score?

2. What is the percentage of the students that scored higher than John?

3. If 50 students were in that class, about how many of them scored lower than John’s score? (You can round your answer to the nearest whole number.)

4. Tom obtained a score of 59. What is Tom’s z score?

5. What is the percentage of students that scored between John and Tom?

6. There are 50 students in the class, so about how many of them would likely score lower than Tom? (You can round your answer to the nearest whole number.)

7. Anna only knows that she scores at 87^th percentile on this exam, what is her z score?

8. Based on the result of the previous question, what would be Anna’s actual score on the exam?

Answer 1

Let X be the exam score

X~N( 65, 52)

a) John's z score = = = 1.8

b) Percentage of students that scored higher than John = P( X> 74) = P( z > 1.8) = 1- P( z<1.8)

= 1- 0.88100

=0.119 = 11.9%

c) Percentage of students who scored lower than john = 1- 0.119 = 0.881 =88.1%

If there are 50 students , then the number of students who scored lower than John's score = 0.881 * 50 = 44.05 =44

d)  Tom’s z score = = = -1.2

e) Percentage of students who scored between John and Tom

= P( 59< X < 74)

= P( -1.2 < z < 1.8)

= P( z < 1.8) - P(z < -1.2)

= P( z < 1.8) - 1 + P (z <1.2)

= 0.88100 - 1 +0.88493

=0.765593

=76.6%

f) Percentage of students sciring lower than Tom = P( X < 59) = P( z < -1.2) = 1- P( z< -1.2) = 1- 0.88493 =0.11507

= 11.5%

There are 50 students,\

Number of students who score lower than tom = 50* 0.11507 = 5.75 = 6

g) Anna scores at 87th percentile

P( Z <z ) =0.87

P( Z < 1.126) = 0.87

So Anna 's z-score = 1.126

h) Anna 's z-score = 1.126

   = 1.126

x = 65 + 1.126*5 = 70.63

Anna actual score is 70.63