Questions 16-18: A company starts a fund of M dollars from which it pays a bonus

Question

Questions 16-18: A company starts a fund of M dollars from which it pays a bonus of $1,000 to each employee who achieves high performance during the year. The probability of each employee achieving this goal is 0.10, and the employees achieve the goal independently of one another. The company has 10 employees. 16. What is the minimum value that M should be so that the fund has 20.99 probability of having enough money to cover the payments? 17. Compute the expected amount of money paid out in bonuses 18. Compute the variance of the amount paid out in bonuses

Explanation / Answer

This is an example of binomial distribution with n=10, p= 0.10 and 1 = 1-p = 0.90

First we will solve 17,18 then 16

17) Expected number of people to get bonus = mean = np = 10*0.10 = 1

This is expected number of people who will get bonus

Expected amount of money paid out = $1000*Expected number of people to get bonus

= $1000*1 = $1000

18) variance in number of people = npq

= 10*0.1*0.9 = 0.9

Thus, variance of the amount paid = $1000*variance in number of people

= $1000*0.9 = $900

16)

std deviation of amount paid = sqrt(variance of amount paid) = sqrt(900) = 30

Thus, range is $(1000-30) to $(1000+30) = $970 to $1030

Thus, miniumum value of funds should be $1030

for probability 0.99 of having enough money, miniumum value of funds should be $(1030*0.99) = $1019.7

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