distributions

[SHIKHA]

thirty percent of all employees have advanced training. if in a sample of 500 employees , less than 27% are adequately prepared, all new hires are required to enroll in a training program. what is the probability the program will be initiated??,,...please tell me your answer

[Amit Goyal]

We are interested in finding:
Pr(less than 27% are adequately prepared (trained) in a sample of 500 given that you are sampling them from a large population with 30% trained)
(Note that This has a Binomial distribution with parameter p = 0.3 and n = 500)
One way to do this is to compute the above probability using Binomial distribution that is, we compute the probability that number of successes is less than or equal to 135 where the draw takes place 500 times (n = 500) and probability of success in each draw is 0.3 (p = 0.3). But doing this is computationally cumbersome.
So we use normal approximation,
Pr[z < (135 - 150)/(sqrt(500(0.3)(0.7)))] = Pr[z < -15/(sqrt(105))] = Pr[z < -15/10.25] = Pr[z < -1.465] = 1 - 0.93 = 0.07
Probability the program will be initiated is approximately 0.07 (7% chance)

[SHIKHA]

thank u sir!!!..