Name: ____________________
Homework #3: Due Friday Sept. 17, 2004
Homework answers should be typed or neatly printed (no cursive). If your answers are not legible,
you will lose points. Use this form as a cover sheet for your homework. Print out these pages, fill
in your name above, and staple the pages with your answers to them.
Remember: For problems that require calculations, SHOW YOUR WORK!
1. A developmental psychologist asked a group of 60 first graders (age 6) to complete a moralreasoning test. The developmental psychologist recorded the kids’ performances on a scale where a
score of 9 - 10 indicates “extremely developed” moral reasoning, 7 - 8 means “moderately
developed” moral reasoning, 5 - 6 means “somewhat developed” moral reasoning, 3 - 4 means
“slightly under developed” moral reasoning, and 1 - 2 means “extremely under developed” moral
reasoning. Below is a frequency distribution of the scores. Use this table to answer the following
questions.
X
8
7
6
5
4
3
2
1
f
1
2
5
11
15
18
5
3
a. What is the probability of a child scoring 6 or above on the test? (1 point)
There are 60 total children. 5 scored 6, 2 scored 7, and 1 scored 8. Therefore the probability
of a child scoring 6 or above on the test = 8/60 = .133.
b. What is the probability of a child scoring 9 on the test? (1 point)
There are 60 total children, and zero of them scored 9 on the test. Therefore the probability
of a child scoring 9 on the test = 0/60 = 0.0.
c. What is the probability of a child scoring in the “somewhat developed” category on the
test? (1 point)
There are 60 total children. A child falls in the “somewhat developed” category if he or she
scores 5 or 6. 11 children scored 5, and 5 children scored 6. Therefore the probability of a
child scoring in the “somewhat developed” category = 16/60 = .267.
2. Based on a large-scale survey of American women, it is known that the average height of US
women is 63.5 inches with a standard deviation of 2.5 inches. Use this information to answer the
following questions.
a. Dr. Claypool is 61 inches tall. Convert her height to a z-score. (1 point)
z=
X 

z = 6163.5  1.0
2.5
b. Susan’s height is represented by a z-score of +1.8. How tall is Susan? (1 point)
X=
 z
X = 63.5  (1.8)(2.5)  68 inches tall
c. What percentage of women are between 60 and 65 inches tall? (1 point)
z=
X 

z = 60 63.5  1.4
2.5
The area below a z-score of –1.40 is equivalent to the “smaller portion” of a z-score of
+1.40, which from Table E.10 is .0808.
z=
X 

z = 6563.5  .60
2.5
The area below a z-score of +.60 is the “larger portion” from this z-score, which from Table
E.10 is .7257.
To find the area between heights of 60 inches (z = -1.40) and 65 inches (z = +0.60), we can
take the smaller value and subtract it from the larger value.
Therefore, the proportion of US women who are between 60 and 65 inches tall is: .7257 .0808 = .6449. Thus, the percentage of US women who are between 60 and 65 inches tall is
64.49%.
d. Angie’s height puts her at the 20th percentile. How tall is Angie? (1 point)
The 20th percentile corresponds to a z-value with .20 “below it.” This means we need a zscore with a smaller portion of .20. Note that this z-score will be NEGATIVE b/c the 20th
percentile is below the mean.
A z-score of .84 is the one closest to having a smaller portion of .20. Thus, the desired zscore is -.84. Now, we need to plug this z-score into the following equation to find the raw
score corresponding to this value:
X=
 z
X = 63.5  (-.84)(2.5)  61.4
Angie is 61.4 inches tall.
e. Suppose that Janet’s height is equivalent to a z-score of +.50. What does this mean?
(Note, I am not asking for a calculation. Just give an interpretation of this z-score). (1
point)
Since Janet’s height results in a positive z-score, we know that her height is above the
average. The magnitude of her z-score (.50) indicates that she is specifically .50 standard
deviations above the average.
3. The 68 students in PSY 293 take their first exam. Dr. Claypool converts each raw exam score to
a z-score. What will be the mean, the standard deviation, and the shape of this z-distribution of
exam scores? (3 points)
The mean of a z-distribution is ALWAYS zero. The standard deviation of a z-distribution is
ALWAYS 1. And, the shape of a z-distribution will be the same shape as that of the raw scores. If
the raw scores were normally distributed, the shape of a z-distribution of those raw scores will also
be normal. If the raw scores were skewed, the shape of a z-distribution of those raw scores will also
be skewed.
4. Greg and Dan are brothers. Greg is in his senior year of college and just took the GRE because
he hopes to attend graduate school. Dan is in his senior year of high school and just took the ACT
because he hopes to attend college. Scores on the GRE Verbal section have  = 500 and  = 20.
Scores on the ACT Verbal section have  = 18 and  = 6. Greg scored 525 on the Verbal GRE, and
Dan scored 23 on the Verbal ACT. Which brother has the better percentile rank? (2 points)
z=
z=
X 

X 

z = 525500  1.25  Greg’s z-score
20
z = 2318  .83 Dan’s z-score
6
From Table E.10, a z-score of 1.25 has an area of .8944 below it. Therefore, Greg is at the 89th
percentile (if rounded). A z-score of .83 has an area of .7967 below it. Thus, Dan’s percentile rank
is 80th (if rounded). Greg performed better on the GRE (i.e., was at a higher percentile rank) than
Dan did on the ACT.
5. A developmental psychologist constructed a questionnaire to assess “psychological maturity”
among 10-year olds. Scores from this measure range from 0 – 56 with a  = 34 and  = 8. The
developmental psychologist has decided to re-scale her measuring instrument so that it has a  =
200 and  = 50. Use this information to answer the following questions.
a. Cliff received a score of 17 (in original units) on the measure. What would Cliff’s score
be in re-scaled ( = 200,  = 50) units? (2 points)
This problem requires us to use transformed standard scores (TSS).
TSS = new + z new
Before we can use the above formula, we must translate Cliff’s score into a z-score.
z=
X 
z = 17 34  2.125

8
Now… TSS = 200 + (-2.125)(50) =93.75
On the re-scaled measure, Cliff scored 93.75
b. Sally’s score on the re-scaled measure is 285. What was her score in the original units?
(2 points)
First, we need to determine what z-score corresponds to this TSS.
TSS = new + z new 285 = 200 + z(50)
If we solve for z we get:
285 - 200 = z(50)
z = 1.70
Now, we need to translate this z-score back to the original units:
X=
 z
X = 34  (1.7)(8)  47.6
Mean from
the original
scale
Thus, Sally’s score in original units is 47.6.
St. dev. from
the original
scale
6. Imagine that you have two bags of marbles. In bag #1, you have 50 blue marbles and 50 yellow
marbles. In bag #2, you have 80 blue marbles and 20 yellow marbles. Suppose you reach into one
of the bags and pull out 10 marbles—3 blue and 7 yellow. From which bag is it more likely that
you obtained your sample and why? [In your answer, be sure to mention how many blue and yellow
marbles you would have expected in your sample had you drawn from Bag #1 and from Bag #2] (3
points)
If you had drawn from Bag #1, you would have expected how many of your 10 to be blue:
Total # of blue marbles in bag: 50
Total # of marbles in the bag: 100
Thus, we would have expected 50% of the marbles (5) to be blue.
If you had drawn from Bag #1, you would have expected how many of your 10 to be yellow:
Total # of yellow marbles in bag: 50
Total # of marbles in the bag:
100
Thus, we would have expected 50% of the marbles (5) to be yellow.
If you had drawn from Bag #2, you would have expected how many of your 10 to be blue:
Total # of blue marbles in bag: 80
Total # of marbles in the bag: 100
Thus, we would have expected 80% of the marbles (8) to be blue.
If you had drawn from Bag #2, you would have expected how many of your 10 to be yellow:
Total # of yellow marbles in bag: 20
Total # of marbles in the bag:
100
Thus, we would have expected 20% of the marbles (2) to be yellow.
To summarize: From Bag #1 we expect: 5 blue and 5 yellow
From Bag #2 we expect: 8 blue and 2 yellow
We actually drew: 3 blue and 7 yellow
The sample we obtained (3 blue; 7 yellow) is much more similar to the expected values from Bag
#1. So, the likelihood is greater that our sample came from bag #1 than bag #2.