Question

The mean per capita income is 18,904 dollars per annum with a standard deviation of 469 dollars per annum. What is the probability that the sample mean would be less than 18850 dollars if a sample of 333 persons is randomly selected? Round your answer to four decimal places.

Answer 1

Solution :

Given that ,

mean = $\mu$ = 18904

standard deviation = $\sigma$ = 469

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 469 / $\sqrt$ 333 = 25.7010

P($\bar x$ < ) = P(($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (18850 - 18904) / 25.7010)

= P(z < -2.1011)

= 0.0178

probability = 0.0178