Question

According to a recent Current population Reports self-employed individuals in the United States work an average of 45 hours per week with a standard deviation of 1.5 hours.

1. Assume that this variable is approximately Normally distributed, what proportion of the population work more than 47 hours per week on average?

b)                     What is the 80^th percentile of this distribution?

c)                     What is the interquartile range of the average hours per week worked by self employed persons?

d)                     Suppose we took a random sample of 75 self-employed people and asked how many hours per week they worked on average. Assume the population standard deviation is 1.5 hours, what is the standard error of the sample mean of a sample of 75 persons?

Answer 1

Solution:- Given that mean = 45, sd = 11.5

a. P(X > 47) = P((X-mean)/sd > (47-45)/11.5)
= P(Z > 0.1739)
= 1 − P(Z < 0.1739)
= 1 − 0.5675
= 0.4325

b. 80th percentile for Z = 0.8416

X = mean + (Z*sd) = 45 + (0.8416*1.5) = 46.2624

c. IQR = Q3 - Q1 = 46.0117 - 43.9883 = 2.0234

Q3 = for Z = 0.6745

Q3 = mean + (Z*sd) = 45 + (0.6745*1.5) = 46.0117

Q1 = for Z = -0.6745

Q1 = mean - (Z*sd) = 45 - (0.6745*1.5) = 43.9883

d) n = 75, sd = 1.5

standard error = sd/sqrt(n) = 1.5/sqrt(75) = 0.1732