Raw scores on behavioral tests are often transformed for easier comparison. A te

Question

Raw scores on behavioral tests are often transformed for easier comparison. A test of reading ability has mean 70 and standard deviation 10 when given to third-graders. Sixth-graders have mean score 80 and standard deviation 11 on the same test. To provide separate “norms” for each grade, we want scores in each grade to have mean 100 and standard deviation 20.

(a) What linear transformation will change third-grade scores x into new scores xnew = a + bx that have the desired mean and standard deviation? (Use b > 0 to preserve the order of the scores.)

(b) Do the same for the sixth-grade scores.

(c) David is a third-grade student who scores 72 on the test. Find David’s transformed score. Nancy is a sixth-grade student who scores 78. What is her transformed score? Who scores higher within his or her grade?

(d) Suppose that the distribution of scores in each grade is Normal. Then both sets of transformed scores have the N(100, 20) distribution. What percent of third-graders have scores less than 75? What percent of sixth-graders have scores less than 75?

Explanation / Answer

a) for  third-grade scores

E(Xnew) =a+b*E(X)

100 =a+70b ..........(1)

and SD(Xnew) =b*SD(X) =20 =b*10

b=2 ; putting it in above: a =100-70*2=-40

therefore transfomed score is = -40+2x

b)

for  sixth-grade scores

E(Xnew) =a+b*E(X)

100 =a+80b ..........(1)

and SD(Xnew) =b*SD(X) =20 =b*11

b=20/11=1.8182 ; putting it in above: a =100-80*1.8182=-45.456

therefore transfomed score is = -45.456+1.8182x

c) David’s transformed score =-40+2*72=104

Nancy   transformed score = -45.456+1.8182*78=96.36

David Scores higher as his standardized transformed score is higher.

d)( here I am calculating percentage based on assuming 75 is the transformed score as mentioned in starting that both sets of transfomred score have the N(100,20) distribution. Please revert if it is not)

percent of third-graders have scores less than 75 =P(X<75)=P(Z<(75-100)/20)=PZ<-1.25)=10.56%

=percent of sixth-graders have scores less than 75 (cause they both have same distribution)

however if we consider them non standardized scores:

percent of third-graders have scores less than 75 =P(X<75)=P(Z<(75-70)/10)=P(Z<0.5)=69.15%

percent of sixth-graders have scores less than 75=P(X<75)=P(Z<(75-80)/11)=P(Z<-0.4545)=32.47%

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