Question

given a percentile rank ((PR) of a score. find the value of score: a) P=72 FOR I SCORES B) P=45 FOR I SCORES. C)P=58 FOR I SCORES D)P=91 FOR IQ SCORES E) P=44 FOR IQ SCORES F) P=17 FOR IQ SCORES G) P=33 FOR IQ H) P=75 FOR GRE SCORES I) P=87 FOR STENS SCORES J) P=37 FOR STENS SCORES

MEANS( X)AND STANDARD DEVIATION (SD)FOR

T -SCALE: (X): (X)= 50; SD=10 IQ- SCALE: (X)= 100; SD= 15 GRE-SCALE: (X)= 500; SD= 100 STENS- SCALE: (X) = 5.5; SD = 2

Answer 1

a)

Let x be The value of score

Percentile Rank is 72 for T score. This means 72% of scores are below x.

P (X<x) = 0.72

There, z = 0.58 (Obtained using z distribution table)

z = (x - x̅)/σ

For T score, x̅ = 50 and σ = 10

0.58 = (x - 50)/10

x = 5.8 + 50 = 55.8

The value of score corresponding to 72 percentile rank is 55.8

b)

Let x be The value of score

Percentile Rank is 45 for T score. This means 45% of scores are below x.

P (X<x) = 0.45

There, z = -0.13 (Obtained using z distribution table)

z = (x - x̅)/σ

For T score, x̅ = 50 and σ = 10

-0.13 = (x - 50)/10

x = -1.3 + 50 = 48.7

The value of score corresponding to 45 percentile rank is 48.7

c)

Let x be The value of score

Percentile Rank is 58 for T score. This means 58% of scores are below x.

P (X<x) = 0.58

There, z = 0.20 (Obtained using z distribution table)

z = (x - x̅)/σ

For T score, x̅ = 50 and σ = 10

0.20 = (x - 50)/10

x = 2 + 50 = 52

The value of score corresponding to 58 percentile rank is 52

d)

Let x be The value of score

Percentile Rank is 91 for T score. This means 91% of scores are below x.

P (X<x) = 0.91

There, z = 1.34 (Obtained using z distribution table)

z = (x - x̅)/σ

For T score, x̅ = 100 and σ = 15

0.58 = (x - 100)/15

x = 8.7 + 100 = 108.7

The value of score corresponding to 91 percentile rank is 108.7

e)

Let x be The value of score

Percentile Rank is 44 for T score. This means 44% of scores are below x.

P (X<x) = 0.44

There, z = -0.15 (Obtained using z distribution table)

z = (x - x̅)/σ

For T score, x̅ = 100 and σ = 15

-0.15 = (x - 100)/15

x = -2.25 + 100 = 97.75

The value of score corresponding to 44 percentile rank is 97.75

f )

Let x be The value of score

Percentile Rank is 17 for T score. This means 17 % of scores are below x.

P (X<x) = 0.17

There, z = -0.95  (Obtained using z distribution table)

z = (x - x̅)/σ

For T score, x̅ = 100 and σ = 15

-0.95 = (x - 100)/15

x = -14.25+ 100 = 85.75

The value of score corresponding to 0.17 percentile rank is 85.75

g)

Let x be The value of score

Percentile Rank is 33 for T score. This means 33% of scores are below x.

P (X<x) = 0.33

There, z = -0.43  (Obtained using z distribution table)

z = (x - x̅)/σ

For T score, x̅ = 100 and σ = 15

-0.43 = (x - 100)/15

x = -6.45 + 100 = 93.55

The value of score corresponding to 33 percentile rank is 93.55

h)

Let x be The value of score

Percentile Rank is 75 for T score. This means 75 % of scores are below x.

P (X<x) = 0.75

There, z = 0.67 (Obtained using z distribution table)

z = (x - x̅)/σ

For T score, x̅ = 500 and σ = 100

0.67 = (x - 500)/100

x = 67 + 500 = 567

The value of score corresponding to 75 percentile rank is 567

i )

Let x be The value of score

Percentile Rank is 87 for T score. This means 87% of scores are below x.

P (X<x) = 0.87

There, z = 1.12  (Obtained using z distribution table)

z = (x - x̅)/σ

For T score, x̅ = 5.5 and σ = 2

1.12 = (x - 5.5)/2

x = 2.24 + 5.5 = 7.74

The value of score corresponding to 87 percentile rank is 7.74

j)

Let x be The value of score

Percentile Rank is 37 for T score. This means 37% of scores are below x.

P (X<x) = 0.37

There, z = -0.33 (Obtained using z distribution table)

z = (x - x̅)/σ

For T score, x̅ = 5.5 and σ = 2

-0.33 = (x - 5.5)/2

x = -0.66 + 5.5 = 4.84

The value of score corresponding to 37 percentile rank is 4.84