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?

***Please explain the steps taken to solve each step, not just an anwser. Thanks!

Explanation / Answer

Pr(Each employye achieving his goal) = 0.10

(a) Here there are n =10 employees then we have to found the value of M so that the fund has a probability of covering atleast 99% of covering those payments.

so, that means let say if there are X employees who acheives high performance.

then,

Pr(x <X ; 10 ; 0.1) >= 0.99

as the given distribution is binomial distribution, we can calculate the cumulative distribution of binomial for which its value is greater than 0.99.

P(X =0) = 10C0 (0.9)10  = 0.3487

P(X =1) =  10C1 (0.9)9 (0.1) = 0.3874

P(X=2) = 10C2 (0.9)8 (0.1)2 = 0.1937

P(X=3) = 10C3 (0.9)7 (0.1)3 = 0.0574

P(X=4) = 10C4 (0.9)6 (0.1)4 = 0.0111

so For Pr(X <= 4) = 0.9983

so M should be equal to 4 * 1000 = $ 4000 to cover 99% of payments.

(b) Here if n = 16 and p=0.1 we have find

Pr(X =2) = Pr(2; 16; 0.1) = 16C2 (0.1)2 (0.9)14 = 0.2745

(c) Here if n = 16 and p = 0.1

Pr(X >=3) = Pr(X >=3 ; 16; 0.1) = 1 - [Pr(X = 0) + Pr(X =1) + Pr(X = 2)]

= 1 - [ 16C0 (0.9)16 (0.1)0 +  16C1 (0.9)15 (0.1) +  16C2 (0.9)14 (0.1)2]

= 1 - 0.7892

= 0.2108

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