Exercise
ABy Admin
May 03'23
Answer
Solution: D
Let [math]X[/math] denote the number of employees who achieve the high performance level. Then [math]X[/math] follows a binomial distribution with parameters n = 20 and p = 0.02. Now we want to determine x such that [math]P[X \gt x] \lt 0.01 [/math] or equivalently
[[math]]
0.99 \leq P[X \leq x ] = \sum_{k=0}^{x} \binom{20}{k}(0.02)^k(0.98)^{20-k}.
[[/math]]
The first three probabilities (at 0, 1, and 2) are 0.668, 0.272, and 0.053. The total is 0.993 and so the smallest [math]x[/math] that has the probability exceed 0.99 is 2. Thus [math]C = 120/2 = 60 [/math].