Question

Question 2. Based on a survey, workers in Ontario earn an average of $60,000 per year with a known standard deviation of $6000. In an attempt to verify this salary level, a random sample of 36 workers in Ontario was selected. Let X represent the mean salary of these 36 workers. a) Describe the sampling distribution of X. b) Calculate the probability that X is between 58,500 and 63,000. c) What is the 90th percentile for X.

Answer 1

Solution :

(a)

The sampling distribution of mean is ,

$\mu$_{$\bar x$} = 60000

The sampling distribution of standard deviation is ,  

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 6000 / $\sqrt$ 36 = 1000

(b)

= P[(58500 - 60000) / 1000 < ($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (63000 - 60000) / 1000)]

= P(-1.5 < Z < 3)

= P(Z < 3) - P(Z < -1.5)

= 0.9318

(c)

P(Z < 1.645) = 0.90

z = 1.645

$\bar x$ = z * $\sigma$ _{$\bar x$}+ $\mu$ _{$\bar x$} = 1.645 * 1000 + 60000 = 61645

90th percentile = 61645