stats hw ch 5 (1)
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School
University of California, San Diego *
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Course
330
Subject
Mathematics
Date
Apr 3, 2024
Type
Pages
4
Uploaded by Anybody1616
Answer the following question using information learned in chapter 5 z-Scores and
location in a distribution. Please show your work. Not showing your work will only earn
you partial credit of 5 points.
1. A set of mathematics exam scores has a mean of 70 and a standard deviation of 8.
A set of English exam scores has a mean of 74 and a standard deviation of 16. For
which exam would a score of 78 have a higher standing?
z(math) = (X - μ) / σ
= (78 - 70) / 8
= 8 / 8
= 1
z(english) = (X - μ) / σ
= (78 - 74) / 16
= 4 / 16
= 0.25
Mathematics has a higher standing
2. In a population of scores a raw score with the value of 83 corresponds to a Z of
+1.00 and a raw score of 86 corresponds to a Z of +2.00. What is the mean and
standard deviation of this population?
1.00=
83−µ
σ
σ= 83-μ
2.00=
86−µ
σ
2.00=
86−µ
83−µ
2(83-
μ)=86-μ
166-2μ=86-μ
-μ=-80
μ=80
σ = 83-μ
σ = 83-80
σ=3
Mean = 80
Standard Deviation = 3
3. For a population with mean of 40 and standard deviation of 8:
Find the z-scores for each of the following X values
X=44
X=48
X=56
X=38
X=34
X=32
Z=
𝑋−µ
σ
X=44
Z=
= 0.5
44−40
8
X=48 Z=
=1
48−40
8
X=56
Z=
= 2
56−40
8
X=38
Z=
= -0.25
38−40
8
X= 34
Z=
= -0.75
34−40
8
X=32
Z=
= -1
32−40
8
4. A score that is 12 points below the mean corresponds to a z-score of z=
-
1.50. what
is the standard deviation?
Z=
𝑋−µ
σ
-1.50 =
−12
σ
x(-1.50) = -12
σ
=
σ
−12
−1.50
= 8
σ
5. A population consists of the following N=5 scores: 2, 4, 1, 7, 2
a. compute the mean and the standard deviation of population
Mean:
μ=
Σ𝑥
𝑁
μ=
= 3.2
2+4+1+7+2
5
Standard Deviation:
=
σ
Σ(𝑋−µ)2
𝑁
=
=
=
=
σ
(2−3.2)
2
+(4−3.2)
2
+(1−3.2)
2
+(7−3.2)
2
+(2−3.2)
2
5
1.44+0.64+4.84+14.44+1.44
5
22.80
5
4. 56
σσ
=2.13
σ
b. find the z-score for each score in the population
Z=
𝑋−µ
σ
X=2
Z=
= -0.56
Z
2= -0.56
2−3.2
2.13
X=4
Z=
= 0.37
Z
4= 0.37
4−3.2
2.13
X=1
Z=
= -1.04
Z
1= -1.04
1−3.2
2.13
X=7
Z=
=1.77
Z
7= 1.77
7−3.2
2.13
X=2
Z=
= -0.56
Z
2= -0.56
2−3.2
2.13
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c. transform the original population into a new population of N=5 scores with mean of 50
and a standard deviation of 10
X
′=
Z
×
σ
′+
μ
′
X=2, Z=-0.56:
X
′= -0.56x10+50=44.4
X=4, Z=0.37:
X
′= 0.37x10+50=53.7
X=1, Z=-1.04:
X
′= -1.04x10+50= 39.6
X=7, Z=1.77:
X
′= 1.77x10+50=67.7
X=2, Z=-0.56:
X
′= -0.56x10+50=44.4
Scores: 44.4, 53.7, 39.6, 67.7, 44.4