Normal Distribution approximation of binomial distribution problem

This problem is meant to be solved using the normal distribution. It is claimed that a bag M&M's contains 24% blue candies. Assuming that this is true, find the probability that at least 27 blue M&M's are in a given package containing 100 total M&M's.

What I did:

n=100, p=.24, q=.76

$\displaystyle \mu =np=(100)(.24)=24$

$\displaystyle \sigma =\sqrt{npq}=\sqrt{(100)(.24)(.76)}=4.2708$

$\displaystyle z = \frac{x-\mu }{\sigma}=\frac{27-24}{4.2708}=.7024$

On a "from the left" z-table I get .7580, so my probability is 1-.7580=.2420

The answer given in the textbook is .2776, or technically .2748 if one uses a binomial distribution. Where did I go wrong? Thanks for taking the time to read this.

Re: Normal Distribution approximation of binomial distribution problem

Hey beebe.

Have you ever come across continuity correction?

Continuity correction - Wikipedia, the free encyclopedia

Re: Normal Distribution approximation of binomial distribution problem

Yes. I'm surprised I forgot about that. Thank you very much.