In: Statistics and Probability
Sophia who took the Graduate Record Examination (GRE) scored 157
on the Verbal Reasoning section and 162 on the Quantitative
Reasoning section. The mean score for Verbal Reasoning section for
all test takers was 150 with a standard deviation of 7.2, and the
mean score for the Quantitative Reasoning was 153 with a standard
deviation of 7.64. Suppose that both distributions are nearly
normal.
(a) What is Sophias Z-score on the Verbal Reasoning section?
On the Quantitative Reasoning section?
(b) Relative to others, which section did she do better on? .
(c) Find her percentile scores for the two exams.
Verbal Reasoning percentile: th
Quantitative Reasoning percentile: th
(d) What percent of the test takers did better than her on the
Verbal Reasoning section? .
On the Quantitative Reasoning section? .
Find the following:
(e) The score of a student who scored in the 75th percentile on the
Quantitative Reasoning section.
(f) The score of a student who scored worse than 85% of the test
takers in the Verbal Reasoning section.
a)
Sophias Z-score on the Verbal Reasoning section
Z =(X - µ ) / σ = ( 157 -
150 ) / 7.2
Z = 0.972
------------------
On the Quantitative Reasoning section
Z =(X - µ ) / σ = ( 162 -
153 ) / 7.64
Z = 1.178
b)
she did better on Quantitative Reasoning section because z score is highest for it
c)
Z =(X - µ ) / σ = ( 157 -
150 ) / 7.2
Z = 0.972
P(X ≤ 157 ) = P(Z ≤
0.97 ) = 0.8345
Verbal Reasoning percentile: 83.45th
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P(X ≤ 162 ) = P(Z ≤
1.18 ) = 0.8806
Quantitative Reasoning percentile: 88.06th
d)
on the Verbal Reasoning sectio = 83.45%.
On the Quantitative Reasoning section = 88.06%
e)
µ = 153
σ = 7.64
proportion= 0.75
Z value at 0.75 =
0.6745 (excel formula =NORMSINV(
0.75 ) )
z=(x-µ)/σ
so, X=zσ+µ= 0.674 *
7.64 + 153
X = 158.153
75th percentile on the Quantitative Reasoning section = 158.15
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f)
µ = 150
σ = 7.2
proportion= 0.85
Z value at 0.85 =
1.0364 (excel formula =NORMSINV(
0.85 ) )
z=(x-µ)/σ
so, X=zσ+µ= 1.036 *
7.2 + 150
X = 157.46(answer)