# Simple Stats Question

I am working on a very simple statistics problem, but I am unsure of two parts of the problem due to unfamiliar wording.

The question is as follows:

Suppose there is a 30% probability that someone has had a flu shot this year. You select 10 people at random.

(a) What is the probability that at least half of these people have had the flu shot?

(b) Find the expected value and standard deviation of the number of those 10 who have received a flu shot.

(c) Find the expected value and standard deviation of the proportion of those 10 who have received a flu shot.

I've completed part (a), simply using $P(x)=\frac{n!} {x!(n-x)!}P^{x}(1-P)^{(n-x)}$ for $P(5), P(6), P(7), P(8), P(9),$and $P(10)$, and then summing the results. The result was $P(5\leq X\leq 10)=0.15042$.

For parts (b) and (c) I am stuck due to not knowing the meaning of the words number and proportion in their context within the question.

My textbook gives the following definitions for the derived mean and variance of a binomial probability distribution.

Let $X$ be the number of successes in $n$ independent trials, each with probability of success $P$. Then $X$ follows a binomial distribution with mean and variance:

$$\mu =E(X)=nP$$

and

$$\sigma_{X}^{2}=E[(X-\mu_{X})^{2}]=nP(1-P)$$

In terms of the question I am answering I would have $P=0.3$ and $n=10$. But there is clearly some other meaning. Thanks for any help.

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I see no problem with what you did in part (b). The word proportion indicates that your random variable is not the number of people vaccinated, but the number of people vaccinated divided by the total number of people. For example, if 6 of the 10 people are vaccinated, $X=0.6$ – Brett Frankel Mar 7 '12 at 17:35
Also, if this is a homework problem, please tag it as such. – Brett Frankel Mar 7 '12 at 17:35

You have in essence given a full answer to (b). The number of people in a sample of $10$ who have had the flu shot is a random variable $X$, with mean $np$ and variance $np(1-p)$, where $p$ is the probability that an individual has had the flu shot.

So the mean of $X$ is $3$, and the variance is $10(0.3)(0.7)$. The standard deviation of the random variable $X$ is the square root of the variance.

For (c), we are looking at a close relative of the random variable $X$. The proportion that received the flu shot is $\dfrac{X}{n}$.

This random variable is frequently denoted by $\overline{X}$. In general, if $X$ is a random variable, and $Y=kX$, where $k$ is a constant, then $$E(Y)=E(kX)=kE(X),$$ and $$\text{Var}(Y)=\text{Var}(kX)=k^2\text{Var}(X).$$ (I am using $\text{Var}$ for variance because it may be easier to read.)

Apply the above formulas, with $Y=\overline{X}=\dfrac{X}{n}$. So $k=\dfrac{1}{n}$. We find that $$E(\overline{X})=\frac{1}{n}E(X)=\frac{1}{n}(np)=p,$$ and $$\text{Var}(\overline{X})=\frac{1}{n^2}\text{Var}(X)=\frac{1}{n^2}np(1-p)=\frac{p(1-p)}{n}.$$

In our situation, we have $p=0.3$, so the mean of the sample proportion, that is, the mean of $\dfrac{X}{n}$, is just $0.3$.

The variance of the sample proportion is $\dfrac{(0.3)(0.7)}{10}$. For the standard deviation, take the square root.

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