Binomial Distribution

#1
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. If there are n = 10 employees,
how much should M equal so that the funds has a probability of at least 99% of covering those
payments?


Is this a binomial distribution question? I am pretty confused. All I could think of is

10
Σ 1000Ck (0.1)^k (0.9)^(1000-k)
k=0

Help pleaseeeeeeee
 

ted00

New Member
#3
cool question

seems you're saying each person is an iid Bernoulli trial
with
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
 

ted00

New Member
#4
also, if this is a real-life question, and there's real $$ to be lost, I'd consider very closely whether the "trials" are truly independent, and I'd probably seek some insight into how sensitive the answer is to the assumed model (which is iid Bernoulli trials, leading to binomial)