Question

In a large population of adults, the mean IQ and standard deviation are

$\mu =112;\sigma =20$

respectively.

Suppose 200 adults are randomly selected for a market research campaign.  What is the probability that the sample mean is less than 110?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 112

standard deviation = $\sigma$ = 20

n = 200

$\mu$$\bar x$ = 112

$\sigma$$\bar x$ =$\sigma$  / $\sqrt$ n = 20/ $\sqrt$ 200=1.4142

P($\bar x$ <110 ) = P[($\bar x$ - $\mu$ $\bar x$ ) / $\sigma$ $\bar x$ < (110 -112) / 1.4142]

= P(z < -1.41)

Using z table  

= 0.0793

probability=0.0793