Question

On an intelligence​ test, the mean number of raw items correct is 233 and the standard deviation is 39. What are the raw​ (actual) scores on the test for people with IQs of​ (a) 120​, ​(b) 87​, and​ (c) 100​? To do this​ problem, first figure the Z score for the particular IQ​ score; then use that Z score to find the raw score. Note that IQ scores have a mean of 100 and a standard deviation of 15. ​(b) What is the raw​ (actual) score on the test for people with an IQ of 87​?

Answer 1

Answer:

Given,

Mean = 233

Standard deviation = 39

a)

First we have to find out the z score & then we calculate the raw score

z score for IQs of 120

z = (X - )/

substitute the values

z = (120 - 233)/39

= -113/39

z = -2.8974

z = - 2.90

Required raw score X = z* +

Here mean = 100

standard deviation = 15

substitute the given values in above expression

X = - 2.90*15 + 100

X = -43.5 + 100

Raw score X = 56.5

b)

Now z score for IQs of 87

z = (X - )/

substitute the values

z = (87 - 233)/39

= -146/39

z = -3.74

Required raw score X = z* +

substitute the given values in above expression

X = -3.74*15 + 100

= -56.1 + 100

Raw score X = 43.9

c)

Now z score for IQs of 100

z = (X - )/

substitute the values

z = (100 - 233)/39

= -133/39

z = -3.41

Required raw score X = z* +

substitute the given values in above expression

X = -3.41*15 + 100

= -51.15 + 100

= 48.85

Raw score X = 48.85