Question

33.     A population consists of the following scores: 12,   1,   10,   3,   7,   3

         A. Compute m and s for the population.

         B. Find the z-score for each raw score in the population.

         C. Transform each score into a new standardized value so that the standardized distribution has a mean of m = 100 and a standard deviation of s = 20.

         Do parts A, B, and C, and then from part C, choose which new score would correspond to the original score of 12:

a.       115

b.      120

c.       130

d.      140

Answer 1

Let us consider the given data can be written as (12,1,10,3,7,3).

A) m=mean of the given data= 36/6=6

s=standard deviation of the given data= sqrt(var)=sqrt(19.2)= 4.38178

B) Z-Score for each raw score in the population

Z-score of 12 is Z=(12-m)/s = (12-6)/4.38= 1.37

Z-score of 1 is Z=(1-m)/s = (12-6)/4.38= -1.41

Z-score of 10 is Z=(10-m)/s = (12-6)/4.38=0.91

Z-score of 3 is Z=(3-m)/s = (12-6)/4.38= -0.68

Z-score of 7 is Z=(7-m)/s = (12-6)/4.38= 0.23

Z-score of 3 is Z=(3-m)/s = (12-6)/4.38= -0.68

C) Transformation of the given data such that mean of standardized variable is 100 & standard devation is 20.

Transformation of the variable is (Z-score * s) + m

For 12, (1.37*20) + 100= 127.4

For 1, (-1.41*20) + 100= 77.18

For 10, (0.91*20) + 100= 118.26

For 3, (-0.68*20) + 100= 86.31

For 7, (0.23*20) + 100= 104.56

For 3, (-0.68*20) + 100= 86.31

new score would correspond to the original score of 12 is 130 since 127.4~ 130.

I have done this problem into R. So, I am attaching my R-code

> x=c(12,1,10,3,7,3)
> m=mean(x)
> s=sd(x)
> z=(x-m)/s
> z
[1] 1.3693064 -1.1410887 0.9128709 -0.6846532 0.2282177 -0.6846532
> m
[1] 6
> s
[1] 4.38178
> z*20 + 100
[1] 127.38613 77.17823 118.25742 86.30694 104.56435 86.30694