Question

General Random Variable

1. A manager in a medium sized company wants to construct an incentive compensation program that equitably and consistently compensates employees on the basis of performance. He decides to offer an annual bonus of $10,000 for superior performance, $6,000 for good performance, $3,000 for fair performance, and no bonus for poor performance. Based on prior performance reviews, he expect 15% of his employees to be superior performers, 25% to be good performers, 40% to be fair performers, and 20% to be poor performers.

What would the probability distribution table for the bonuses look like, if this year’s performance of employees is comparable to the past performance?

What is the chance a randomly selected employee would get a non-zero bonus amount?

What is the chance a randomly selected employee would get at least $6000?

What is the expected bonus value of the bonus amounts?

What are the variance and standard deviation of the bonus amount?

Answer 1

X : Annual Bonus

P : Performance

P =SP : Superior perrformance : SP X : Annual bonus = $10,000 ; 15% of his employees to superior perfomance

P(SP) = Probability of a employee to be superior performer = 15/100 = 0.15

P(SP) = P(X=10000) = 0.15

Similarly ,

GP : Good performance ; Annual bonus = $6,000 ; 25% of his employees to good perfomance

P(SP) = P(X=6000) = 0.25

FP : Fair performance ;  Annual bonus = $3,000 ; 40% of his employees to fair perfomance

P(FP) = P(X=3000) = 0.40

PP : Poor performance = No bonus ; 20% of his employees to poor perfomance

P(FP) = P(X=0) = 0.20

probability distribution table for the bonuses

X: Bounus	10,000	6,000	3000	0
P(x)	0.15	0.25	0.40	0.20

Chance a randomly selected employee would get a non-zero bonus amount = P(X>0) = 1 - P(X0) = 1-P(X=0) = 1-0.20 = 0.80

Chance a randomly selected employee would get a non-zero bonus amount = 0.80

chance a randomly selected employee would get at least $6000 = P(X6000) = P(X=6000) + P(X=10000) = 0.25+0,15 = 0.40

Chance a randomly selected employee would get at least $6000 = 0.40

expected bonus value of the bonus amounts = E(X)

X: Bounus	10,000	6,000	3000	0	Total
P(x)	0.15	0.25	0.40	0.20	1
x.P(x)	10000x0.15=1500	6000x0.25=1500	3000x0.40=1200	0x0.20=0
x.P(x)	1500	1500	1200	0	1500+1500+1200+0=4200

variance of the bonus amount: Var(X)

Var(X) = E(X²) - E(X)²

X: Bounus	10,000	6,000	3000	0	Total
P(x)	0.15	0.25	0.40	0.20
x.P(x)	1500	1500	1200	0	4200
x2.P(x)	15000000	9000000	3600000	0	27600000

Var(X) = E(X²) - E(X)² = 27600000 - 4200² =  27600000 - 17640000=9960000

Var(X) = 9960000

standard deviation of the bonus amount:

variance of the bonus amount =  9960000

Standard deviation of the bonus amount = 3155.9468