A binomial distribution has 100 trials n 100 with a probabi

A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 (?=0.25). To apply the normal distribution to approximate the binomial, what are the mean and standard deviation?

Solution

mean=n*p=100*0.25 =25

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

=sqrt(100*0.25*0.75)

=4.330127

Answer: