Here is necessary background information for the remaining five questions on thi

Question

Here is necessary background information for the remaining five questions on this problem set: According to the U.S. Census Bureau, the population distribution of number of years of education for self-employed individuals in the United States in 1990 had a mean of 13.6 and a standard deviation of 3.0 8. Regarding this population: Assuming the distribution is normal, what can you say about the proportion of these individuals with between 9.1 and 18.1 years of education? The least educated 20 percent of the self-employed have below how many years of education? For this part, also assume the distribution is normal a) b)

Explanation / Answer

Question 8

Here mean years of education for self employed individuals = 13.6 years

Standard deviation = 3.0 years

(a) Here if x is the number of years of education an individual posses.

Pr(9.1 < x < 18.1) = Pr(x < 18.1) - Pr(X < 9.1) = NORM (x < 18.1 ; 13.6 ; 3.0) - NORM (x < 9.1 ; 13.6 ; 3.0)

Z2  = (9.1 - 13.6)/3.0 = -1.5 ; Z1  = (18.1 - 13.6)/3.0 = 1.5

Pr(9.1 < x < 18.1) = NORM (x < 18.1 ; 13.6 ; 3.0) - NORM (x < 9.1 ; 13.6 ; 3.0) = Pr(Z < 1.5) - Pr(Z < -1.5) = 0.9332 - 0.0668 = 0.8664

(b) Here the least 20 percent of the self employed have lets say C years of education

so Pr(x < C) = 0.20

as the distribution is normal.

Pr(x < C ; 13.6 ; 3.0) = 0.20

Z value = -0.8416

(C - 13.6)/3 = -0.8416

C = 13.6 - 3 * 0.8416 = 11.075 year

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