MHB How to Compute Standard Deviation in a Mixed Sampling Problem

[Ackbach]

This is problem AP3.7 on page 669 of The Practice of Statistics, 5th AP Ed., by Starnes, Tabor, Yates, and Moore.

A certain candy has different wrappers for various holidays. During Holiday 1, the candy wrappers are 30% silver, 30% red, and 40% pink. During Holiday 2, the wrappers are 50% silver and 50% blue. Forty pieces of candy are randomly selected from the Holiday 1 distribution, and 40 pieces are randomly selected from the Holiday 2 distribution. What are the expected value and standard deviation of the total number of silver wrappers?

Now, I've computed the expected value of silver candies as $40(0.3)+40(0.5)=32$. But I am at a loss to compute the standard deviation. My instinct tells me this is a discrete random variable, in which case I should compute
$$\sigma=\sqrt{\sum_i(x_i-\overline{x})^2 p_i}.$$
But then what are the $x_i$ and $p_i$ values?

 

[Jameson]

I would think of this as a binomial random variable, since we essentially have two outcomes and are considering each draw to be independent of the rest. For Holiday 1, each draw has a 30% "success" rate, thus a 70% "failure" rate. Same idea for Holiday 2. You found the expected value correctly using the formula for a binomial distribution $E[X]=np$. I think you can continue in this way by using the formula for the variance of a binomial random variable: $np(1-p)$. Also, since Holiday 1 and Holiday 2 are assumed to be independent, we can simply add their variances together to get the combined variance, then find the standard deviation.

 

[I like Serena]

Hi Ackbach,

The random variable $X$ is the number of silver wrappers.
The possible outcomes are $x_1=0$ up to $x_{81}=80$ silver wrappers with a population size of $n=81$.
The $p_i$ are the corresponding probabilities and for instance $p_{81} = P(80 \text{ silver wrappers}) = 0.3^{40} \cdot 0.5^{40}$.

Since this is about a population instead of a sample the proper symbol for the mean is $\mu$ instead of $\bar x$.
$$\sigma = \sqrt{\sum(x_i - \mu)^2 p_i}$$

As Jameson said, this is the sum of 2 binomial distributions.

When summing distributions, we have:
\begin{cases}
\mu_{_{Y+Z}} &= \mu_{_Y} + \mu_{_Z} \\
\sigma_{_{Y+Z}}^2 &=\sigma_{_Y}^2 + \sigma_{_Z}^2 + 2\sigma_{_Y}\sigma_{_Z}\rho_{_{YZ}}
\end{cases}where $\rho_{_{YZ}}$ is the correlation between $Y$ and $Z$.

 

[Ackbach]

Thanks very much for your replies, Jameson and I like Serena. They cleared things up immensely in my mind! I was able to obtain the correct answer.

Cheers.