Question

computing z scores

A researcher wanted to study the effects of mentoring on intelligence scores. He wanted to know as a baseline what the average intelligence of his students were relative to the general population. He used a standardized IQ test which has a mean of 100 and standard deviation of 16. The 50 students in his study scored an average of 102 on the IQ test. At what percentile of intelligences is the average IQ of his students relative to his population?

Answer 1

SOLUTION:

Solution :

Given that ,

mean = $\mu$ = 100

standard deviation = $\sigma$ = 16

n=50

$\mu$_{$\bar x$} = 100

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 16 / $\sqrt$ 100 = 1.6

USING Z SCORE FORMULA

Z=($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$}

=(102 - 100 ) / 1.6

z=1.25

p(z<1.25)=0.8944

=89.44% percentile of intelligences is the average IQ of his students relative to his population

Answer 2

=89.44% percentile of intelligences is the average IQ of his students relative to his population