Question

the mean per capita income.is 21604 dollars per annum with a standard deviation of 727. what is the probability.that the sample.mean. wouldnsiffer from the true mean by more than 36 dollars.if a sample of 193 persons is randomly.selected

Answer 1

Solution :

Given that,

mean = $\mu$ = 21604

standard deviation = $\sigma$ = 727

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 727 / $\sqrt$ 193

= 1 - P[-36 / 727 / $\sqrt$ 193 < ($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < 36 / 727 / $\sqrt$ 193]

= 1 - P(-0.69 < Z < 0.69)

= 1 - P(Z < 0.69) - P(Z < -0.69)

= 0.4902