Question

The mean number of establishments responding to the Bureau of Labor Statistics (BLS) Current Employment Statistics survey is 450,000 each month. The BLS selects a random sample of 12 months to monitor the month-to-month change in the number of survey respondents. Assume the population standard deviation for monthly respondents is σ = 24,580, the monthly number of respondents is normally distributed, and this sample of 12 months is small relative to the size of the population.

1. What is the probability that the sample mean will be within 5,000 establishments of the population mean?
2. What is the probability that the sample mean will be within 12,500 establishments of the population mean?

Answer 1

Solution :

Given that,

mean = $\mu$ = 450000

standard deviation = $\sigma$ = 24580

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 24580 / $\sqrt$ 12 = 7095.6348

a.

= P[(445000 - 450000) / 7095.6348 < ($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (455000 - 450000) / 7095.6348)]

= P(-0.70 < Z < 0.70)

= P(Z < 0.70) - P(Z < -0.70)

= 0.758 - 0.242

= 0.516

Probability = 0.516

b.

= P[(437500 - 450000) / 7095.6348 < ($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (462500 - 450000) / 7095.6348)]

= P(-1.76 < Z < 1.76)

= P(Z < 1.76) - P(Z < -1.76)

= 0.9608 - 0.0392

= 0.9216

Probability = 0.9216