Question

Assume that IQ scores are normally distributed, with a standard deviation of 12 points and a mean of 100 points. If 50 people are chosen at random, what is the probability that the sample mean of IQ scores will not differ from the population mean by more than 2 points?

(Round your answer to four decimal places.)

Answer 1

Solution :

Given that ,

mean = $\mu$ = 100

standard deviation = $\sigma$ = 12

n = 50

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

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 12 / $\sqrt$ 50 = 1.6971

We have find the probability that the sample mean of IQ scores will not differ from the population mean by more than 2 points.

P(98 < $\bar x$ < 102) = P((98 - 100) / 1.6971<($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (102 - 100) /1.6971 ))

= P(-1.18 < Z < 1.18)

= P(Z < 1.18) - P(Z < -1.18) Using standard normal table,  

= 0.8810 - 0.1190

= 0.7620

Probability = 0.7620