Question

A company establishes a fund of 120 from which it wants to pay an amount,C, to any of its 20 employees who achieve a high-performance level during the coming year. Each employee has a 2% chance of achieving a high-performance level during the coming year, independent of any other employee.
Determine the maximum value of C for which the probability is less than 1% that the fund will be inadequate to cover all payments for high performance.

Answer 1

Answer:

$C=120/2=60$

Step by step Explanation'

To solve this problem, we will need to apply trial-and-error calculation with the binomial distribution, even though it appears like Central Limit Theorem but it's not.

For us to know the value of C , we will look for a minimum integer such that having 'n' number of high performance level of employee has the probability below 0.01.

Determine the maximum value of C, then the maximum value that C can have is 120/n

Let us represent X as the number of employees with high performance with a binomial distribution of

P =0.02( since the percentage of chance of achieving a high performance level is 2%)

n = 20 ( number of employees who achieve a high performance level)

The probability of X= 0 can be calculated

P( X= 0) = 0.98^n

$P(X=0)=0.98^20$

$P(X=0)=0.668$

$P(X=1)=0.02*20*0.98^19$

$P(X=1)=0.272$

$P(X=2)=0.02^2*20*0.98^18$

$P(X=2)=0.053$

Summation of P( X= 0)+ P( X= 1)+P( X= 2) will give us the value of 0.993 which is greater than 0.99( 1% that the fund will be inadequate to cover all payments for high performance.)

BUT the summation of P( X= 0)+ P( X= 1) will give the value of 0.94 which doesn't exceed the 0.99 value,

Therefore, the minimum value of integer in such a way that P(X >2) is less than 0.01 have n= 2

then the maximum value that C can have is 120/n

$C=120/2=60$