Can someone please help...Mars, Inc. claims that 24% of its M&M plain candies are blue. A sample of 100 M&Ms is randomly selected. Find the mean and standard deviations for the number of blue M&Ms in such a group of randomly selected 100 candies. This is what I have so far: Mean = np = 100(.24) = 24 Stdev = 100 x .24 x

Question

kropot72 · Answer

The standard deviation for a binomial distribution is found as follows:
$$s.d.=\sqrt{np(1-p)}$$$$=\sqrt{100\times0.24(1-0.24)}=you\ can\ calculate$$

kropot72 · Answer

@brettwlvn Do you understand?

anonymous · Answer

i took a minute to eat something...sorry. I didn't think I had gone far enough.

anonymous · Answer

Were my calculations correct?

kropot72 · Answer

Your calculation for the expected mean in the random sample of 100 is correct.

anonymous · Answer

Ok, thanks. I'm going back to school at the age of 50; it's been some time since my last math class!  =]

kropot72 · Answer

You're welcome :) Good wishes!

anonymous · Answer

Might you have a moment to help me work this through to the end?

anonymous · Answer

is it...   Stdev = 100 x .24 x .76 = 18.24  ?

kropot72 · Answer

Not really. Did you notice the radical symbol, which means you need to take the square root of 18.24 to find the standard deviation.
Note: 18.24 is the variance, which is not requested.