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Raw scores on behavioral tests are often transformed for easier comparison. A test of reading ability has mean 67 and standard deviation 10 when given to third graders. Sixth graders have mean score 86 and standard deviation 9 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. (Round your answers to two decimal places.)
(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.) a = b = (b)
Do the same for the sixth-grade scores.
a =
b =
(c) David is a third-grade student who scores 79 on the test. Find David's transformed score.
Nancy is a sixth-grade student who scores 79. What is her transformed score?
Who scores higher within his or her grade? Nancy or David

Sagot :

ogorwyne

Answer:

a.) a = -34 and b = 2

b.) a = -91.11 b = 2.22

c.) David scored 124, nancy scored 84.27

David scored higher

Step-by-step explanation:

for those in the 3rd grade

mean u = 67

s = 10

we used the z formula

[tex]z=\frac{x-u}{s}[/tex]

[tex]z=\frac{x-67}{10}[/tex]

we have the new score

u = 100

s = 20

we use this formula [tex]z*s+u[/tex]

[tex][\frac{x-67}{10} ]20\\[/tex] + 100

multiply through by 20

[tex][\frac{20x-1340}{10} ] +100[/tex]

dividing the bracket by 10

= [tex]2x-134+100[/tex]

= [tex]2x-34[/tex]

from this solution above,

a = -34 and b = 2

b.)

we solve for those in the 6th grade

u = 86

s = 9

[tex]\frac{x-86}{9}[/tex]

for new score,

y = 100

s = 20

[tex][\frac{x-86}{9} ]20+100[/tex]

[tex]2.22x-191.11+100\\= 2.22x - 91.11[/tex]

from this solution

a = -91.11

b = 2.22

c.) if David scored 79

[tex]= 2(x)-34[/tex]

= 2(79) - 34

= 158 - 34

= 124

if nancy scored 79

= 2.22(79) - 91.11

= 175.38 - 91.11

= 84.27

from these scores we can see that David scored higher in his grade

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