can any1 help me out....A random number generator draws at random with replacement from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. In 5000 draws, the chance that the digit 0 appears fewer than 495 times is closest to
a)25% b)30% c) 35% d) 40% e) 45% f) 50%

Question

can any1 help me out....A random number generator draws at random with replacement from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. In 5000 draws, the chance that the digit 0 appears fewer than 495 times is closest to
a)25%  b)30%  c) 35%  d) 40%  e) 45%  f) 50%

zarkon · Answer

This is a binomial distribution problem

anonymous · Answer

@agent0smith

agent0smith · Answer

Normal approximation to binomial

p = 0.1 of getting a zero
mean E(x) = np = 0.1*5000 = 500
SD = sqrt(np(1-p)) = sqrt(500*0.9) = 21.21


for normal approximation
z = (494.5-500)/21.21 = -0.26

P(x < 495) = P(X < 494.5) = 39.7%

anonymous · Answer

got it...:)...thanks dear..