A company starts a fund of M dollars from which it pays $1000 to each employee w

Question

A company starts a fund of M dollars from which it pays $1000 to each employee who achieves high performance during the year. The probability of each employee achieving this goal is 0.10 and is independent of the probabilities of the other employees doing so. (a) If there are n = 10 employees, how much should M equal so that the fund has a probability of at least 99% of covering those payments? (b) If there are n = 16 employees, what is the probability that there are only two employees who achieve high performance? (c) If there are n = 16 employees, what is the pr obability that there are three or more employees who achieve high performance?

Explanation / Answer

a) N=10
p=0.10
x=number "successes"
the binomial distribution is (x=0,...,10)
0.3486784 0.7360989 0.9298092 0.9872048 0.9983651
0.9998531 0.9999909 0.9999996 1.0000000 1.0000000 1.0000000
So,
Pr(x<=3)=0.987
Pr(x<=4)=0.998
meaning (if I'm understanding correctly) M=$4000

b) P = 16C2 * 0.1^2 * 0.9^14

P = 0.2745

c) 3 or more

P = 0.2108

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