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

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

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 =
`(\bar X -\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

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

`\sigma` = standard deviation =
`√(Variance)` =
`√(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(
`(\bar X -\mu)/((\sigma)/(√(n) ) )` <
`(23,013-23,037)/((387)/(√(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.