Question

The mean per capita income is 23,037 dollars per annum with a variance of 149,769. What is the probability that the sample mean would be less than 23013 dollars if a sample of 134 persons is randomly selected? Round your answer to four decimal places.

Answer 1

Answer:

Probability that the sample mean would be less than 23,013 dollars is 0.2358.

Step-by-step explanation:

We are given that the mean per capita income is 23,037 dollars per annum with a variance of 149,769.

Also, a sample of 134 persons is randomly selected.

Firstly, Let $\bar X$ = sample mean

The z-score probability distribution for sample mean is given by;

              Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean per capita income = 23,037 dollars p.a.

           $\sigma$ = standard deviation = $\sqrt{Variance}$ = $\sqrt{149,769}$ = 387 dollars p.a

           n = sample of persons = 134

So, probability that the sample mean would be less than 23,013 dollars is given by = P($\bar X$ < 23,013 dollars)

      P($\bar X$ < 23,013) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{23,013-23,037}{\frac{387}{\sqrt{134} } }$ ) = P(Z < -0.72) = 1 - P(Z $\leq$ 0.72)

                                                                    = 1 - 0.7642 = 0.2358

The above probability is calculated using the z-score table.

Therefore, probability that the sample mean would be less than 23013 dollars if a sample of 134 persons is randomly selected is 0.2358.