Suppose that Y is the sum of X1 X2 Xn which are iid Bernoul

Suppose that Y is the sum of X₁, X₂, ..., X_n, which are i.i.d. Bernoulli random variables with p=.5 and n=400. In other words, Y is binomial with n=400 and p=5. Use Table 4 and the normal approximation to the binomial distribution with the continuity correction to approximate P(Y>>210).

Note: >> = Y is greater than or equal to 210.

Solution

mean=n*p=400*0.5 = 200

standard deviation =sqrt(n*p*(1-p))

=sqrt(400*0.5*0.5)

=10

So the probability is

P(Y>=210) = P(Y>209.5) (by using continuity correction)

=P((Y-mean)/s >(209.5-200)/10)

=P(Z>0.95) = 0.1711 (from standard normal table)