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View Full Version : Exam #1, Novmber 2001, Question #27

01-10-2002, 04:18 PM
This one has me stumped:

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.

The answer key says C = 60.

Using the pivotal method with the normal approximation gives C = 64.64.

Since n is small, and the density function is skewed, this is not the best approach, but what other method is available in the middle of an exam?

Fortal
01-10-2002, 04:35 PM
On 2002-01-10 16:18, Sladewski wrote:
This one has me stumped:

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.

The answer key says C = 60.

Using the pivotal method with the normal approximation gives C = 64.64.

Since n is small, and the density function is skewed, this is not the best approach, but what other method is available in the middle of an exam?

When a problem like this comes, with n not large enough for the normal approximation and with very low probabilities involved, you should suspect that even though in principle it is very hard to do it exactly, what the problem is actually asking is not that hard to get. So one should solve it exactly, using the binomial distribution:

We want to find n (# of High Performance Employees) such that
Pr(n C < 120) < 1%
The probability that none of the employees is HP is (.98)^20 = .6676
Similarly, one will be HP with probability
20 (.98)^19 .02 = .2725
And two will be HP with probability
20 19 / 2 (.98)^18 (.02)^2 = .0528
and the numbers keep getting smaller. Note that the sum of these three numbers is already larger than 99%, so that the number n we are looking for is 2, and C=120/2=60.

Fortal
01-10-2002, 04:54 PM
Sorry, there's a typo in my solution:
The expression should read

Pr(n C > 120) < 1%

since the event is the fund (120) not being enough to cover costs (n times C).