Business, 02.01.2020 23:31 KadaLearns

In the workplace, racial discrimination is a very serious issue. consider a company in which 20% of the employees are african-american. at the end of the year, promotions are awarded to a group of employees. out of the 40 promotions awarded, five are african-american. given that the awarding follows the binomial distribution, b(40,.2).

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Answer from: gabbyallen9703

a) It is expected that 8 African-Americans get promotions.

b) There is a 8.6% probability that 5 African-Americans get promotions.

c) There is a 16.2% probability that at five or less African-Americans get promotions.

d) The company may be accused of racial discrimination because the ammount of promotions given to African-Americans is much less than expected if there were no discrimination. The expected value, if there is no discrimination, of having more than 5 promotions for African American employees is 84%.

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Complete question:

In the workplace, racial discrimination is a very serious issue. Consider a company in which 20% of the employees are African-American. At the end of the year, promotions are awarded to a group of employees. Out of the 40 promotions awarded, five are African-American. Given that the awarding follows the binomial distribution, B(40,.2).

a) How many African-Americans would you expect to get promotions?

b) What is the probability that five African-Americans receive promotions?

c) What is the probability that five or fewer African-Americans receive promotions?

d) Do you think the company is suspect of racial discrimination? Explain your thinking.

a) As this situation can be modeled by a binomial distribution B(40,0.2), the expected number of African-Americans that get promotions can be calculated as the expected value of the binomial distribution:

$X\sim B(40,0.2)\\\\E(X)=np=40*0.2=8$

It is expected that 8 African-Americans get promotions.

b) Accordingly to the binomial distribution, we have:

$P(X=5)=\frac{40!}{5!35!}*(0.2)^5*(0.8)^{35}=658008*0.00032*0.0004= 0.086$

There is a 8.6% probability that 5 African-Americans get promotions.

c) We have to calculate the probabilities for X=0,1,2,3,4 and 5.

$P(X\leq5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\\\P(X=0)=\frac{40!}{0!40!}*0.2^0*0.8^{40}=1*1*0.00013=0\\\\P(X=1)=\frac{40!}{1!39!}*0.2^1*0.8^{39}=40*0.2*0.00017=0.001\\\\P(X=2)=\frac{40!}{2!38!}*0.2^2*0.8^{38}=780*0.04*0.00021=0.007\\\\P(X=3)=\frac{40!}{3!37!}*0.2^3*0.8^{37}=9880*0.008*0.00026=0.021\\\\P(X=4)=\frac{40!}{4!36!}*0.2^4*0.8^{36}=91390*0.0016*0.00032=0.047\\\\P(X=5)=\frac{40!}{5!35!}*0.2^5*0.8^{35}=658008*0.00032*0.00041=0.086$

$P(X\leq5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=0+0.001+0.007+0.021+0.047+0.086=0.162$

There is a 16.2% probability that at five or less African-Americans get promotions.

Answer from: Quest

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Answer from: Quest

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