Question

A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ test, what is the probability that the sample mean scores will between 102 and 110 points? (Round to 2 decimal places)

Answer 1

Solution :

Given that ,

mean = $\mu$ = 105

standard deviation = $\sigma$ = 20

n = 20

$\mu$_{$\bar x$} = 105

$\sigma$_{$\bar x$} =$\sigma$  / $\sqrt$ n = 20/ $\sqrt$ 20=4.47

P(102< $\bar x$   < 110) = P[(102-105) / 4.47< ($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$}  < (110-105) / 4.47)]

= P( -0.67< Z <1.12 )

= P(Z < 1.12) - P(Z < -0.67)

Using z table,  

= 0.8686-0.2514

=0.6172

=0.62