Question

The mean per capita income is 19,292 dollars per annum with a variance of 540,225. What is the probability that the sample mean would be less than 19269 dollars if a sample of 499 persons is randomly selected? Round your answer to four decimal places.

Answer 1

The probability is 0.2423.

Explanation:

Given mean per capita = 19292 dollars

Given the variance = 540225

Now find the probability that the sample mean will be less than 19269 dollar when the sample is 499.

Below is the calculation:

`\bar{X} \sim N(\mu =19292, \ \sigma = (√(540225))/(√(499))) \\\bar{X} \sim N(\mu =19292, \ \sigma = 32.90) \\\text{therefore the probability is:} \\P (\bar{X}< 19269) \\\text{Convert it to standard normal variable.} \\P(Z< (19269-19292)/(32.90)) \\P(Z< - 0.6990) \\\text{Now getting the probability from standard normal table}\\P(Z< -0.6990) = 0.2423`