DON'T KNOW HOW TO DO THIS!!!!

Question

A candy company claims that 16% of its plain candies are orange, and a sample of 200 such candies is randomly selected.
a. Find the mean and standard deviation for the number of orange candies in such groups of 200.

μ=
σ=(Round to one decimal place as needed.)

Answer 1

First it is important to note the type of distribution such sampling implies.  When choosing a candy, it can either be "orange" or "not orange".  This is binary in nature.  In addition, the size of the sample "n", is fixed at 200 each time the sampling is done.  (We don't know what the population size is but we can assume that it is much, much larger than 200. We can either assume sampling with replacement or that the population size is so large that sampling without replacement will not significantly affect the outcome.)  Thus a certain proportion, "p", of candies will be orange and the remaining proportion, "1-p", is not orange.  This describes a Binomial Distribution (c.f., https://en.wikipedia.org/wiki/Binomial_distribution ).  By formula, but also by intuitive reasoning, if a proportion p are orange, then n*p should give the mean number of orange candies to be found in a random drawing of 200.  Thus the mean is 200*0.16 = 32.  Next, the variance of a binomial distribution is n*p*(1-p) = 200*0.16*0.84 = 26.88.  The standard deviation is just the square root of the variance, thus the standard deviation = sqrt(26.88) = 5.18.

 

John C.

Answer 2

Rodger,

 

This is a binomial distribution with a sample size n and probability p. In this case, n=200 and p=.16. According to the binomial distribution, the mean is np or 32 and the standard deviation is square root of npq where q=1-p:

sqrt(200*.16*.84)=5.18=5.2 in this case.

 

 

Hope that helps!