Binomial Distribution

Question

Suppose a certain experiment yields a binomial distribution with n = 64 trials and a probability of success of p = 0.409. What is the probability of observing at most 27 successes out of the 64 trials? Round your answer to three decimal places; add trailing zeros as needed.

The probability of observing at most 27 successes out of 64 trials is 

Answer 1

So X follows Binomial with n=64,p=.409

Mean=np=64*.409=26.176 and Variance=n.p(1-p)=15.470

you have to find the probability P(X<=27)

Now by CLT, Z=(X-mean)/Standard deviation follows standard Normal distribution.

So, P(X<=27)=P[Z<=(27-26.176)/3.933 ]= P[Z<=.2095] =.583 (Can be found from any standard Normal Calculator)