Question

# Given a binomial random variable with n = 48 and p = 0.62 find the probability...

Given a binomial random variable with n = 48 and p = 0.62 find the probability of obtaining more than 21 but no more than 34 successes to three decimal places.

Probability of success = .62

Probability of failure = 1-.62 = .38

Required probability = (48C22*.62^22 * .38^26 + 48C23*.62^23 * .38^25 + 48C24*.62^24 * .38^24 + 48C25*.62^25 * .38^23 + 48C26*.62^26 * .38^22 + 48C27*.62^27 * .38^21 + 48C28*.62^28 * .38^20 + 48C29*.62^29 * .38^19+ 48C30*.62^30 * .38^18 + 48C31*.62^31 * .38^17 + 48C32*.62^32 * .38^16 + 48C33*.62^33 * .38^15 + 48C34*.62^34 * .38^14

Required probability = 27385657281648*.62^22 * .38^26 + 30957699535776 * .62^23 * .38^25 + 32247603683100*.62^24 * .38^24 + 30957699535776*.62^25 * .38^23 + 27385657281648*.62^26 * .38^22 + 22314239266528*.62^27 * .38^21 + 16735679449896*.62^28 * .38^20 + 11541847896480*.62^29 * .38^19+ 7309837001104*.62^30 * .38^18 + 4244421484512*.62^31 * .38^17 + 2254848913647*.62^32 * .38^16 + 1093260079344*.62^33 * .38^15 + 482320623240*.62^34 * .38^14

Required probability = .915