Help please!
Suppose that the number of children in U.S. families is Binomially distributed with parameters $m=10$ and $p=0.29$. A random sample of $50$ U.S. families is taken.
1. What is the distribution of the sample mean?
I don't know if it was an error and instead of $m=10$ should be $n=10$ or my $n$ is $50$? I'm confused. Anyway, if $n=10$, then since $n \ngtr 30$ and $np=2.9 \ngtr10$ and $n(1-p)=7.1 \ngtr 10$ I can't use normal approximation. So then the distribution of the sample mean is just $X$~$B(10,0.29)$?
Is this ok, or do I have to do the checking of $np>10$ and $n(1-p)>10$ using $n=50$? Because if I use $n=50$ then $np>10$ and $n(1-p)>10$ and then I can use normal approximation. Therefore, the distribution of the sample mean is
$X$~$N(\mu_\hat{p},\sigma_\hat{p})$
$X$~$N(p,\sqrt{\frac{p(1-p)}{n}})$
$X$~$N(0.29,\sqrt{\frac{0.29(1-0.29)}{10}})$
$X$~$N(0.29,0.143492)$
This is confusing to me because I've seen before Binomial Distribution associated with the distribution of the sample proportion not with the distribution of the sample mean.
A single randomly sampled family has a number of children that is binomially distributed with $m = 10$ and $p = 0.29$, as the problem states.
However, for a sample of $n = 50$ families, the mean number of children per family will not in general be an integer, so we can immediately exclude a binomial distribution. For instance, I can generate an example:
$$\{2,3,4,3,4,4,1,3,2,2,4,5,4,2,1,0,5,4,2,4,2,3,3,3,3, \\ 0,1,2,3, 0,2,1,3,1,5,4,3,4,3,4,5,2,4,3,3,2,3,2,5,3\}$$
and the sample mean is $141/50 = 2.82$, which of course is noninteger.
However, the sample mean is also not a continuous random variable, since we know that, being some fraction of $50$ families, it is impossible for it to be an irrational number. So the exact distribution is also not normal.
We can take one of two approaches: reinterpret the question to mean that it is asking for the approximate distribution of the sample mean; or state the exact distribution in terms of a probability mass function.
The first approach, assume the problem wants an approximate distribution, implies that we will use the normal approximation for a sufficiently large (in this case $n = 50$) sample size. Then such an approximation uses the binomial mean and variance, which are $\mu = mp = 2.9$ and $\sigma^2 = mp(1-p) = 2.059$, respectively. So the mean and variance of the sample mean will be $$\mu = 2.9, \quad \frac{\sigma^2}{n} = 0.04118,$$ hence the sample mean is approximately normally distributed with mean $2.9$ and variance $0.04118$.
The second approach is to give the exact PMF of the sample mean: The relevant variable here is $$\bar X = \frac{1}{50} (X_1 + X_2 + \cdots + X_{50})$$ where $X_i \sim \operatorname{Binomial}(m = 10, p = 0.29)$ for $i = 1, 2, \ldots, 50$, so the sample total is $$50 \bar X \sim \operatorname(mn = 500, p = 0.29).$$ That is to say, the total number of children across the sample of $50$ families, is a binomial random variable whose support ranges from $0$ to $500$ children, and the Bernoulli probability parameter of a child remains $0.29$. So $50\bar X$ has PMF $$\Pr[50 \bar X = x] = \binom{500}{x} (0.29)^x (0.71)^{500 - x}, \quad x \in \{0, 1, \ldots, 500\},$$ but the PMF of $\bar X$ itself will need to be written as $$\Pr[\bar X = k] = \binom{500}{50k} (0.29)^{50k} (0.71)^{500 - 50k}, \quad k \in \{0, \tfrac{1}{50}, \tfrac{2}{50}, \ldots, \tfrac{49}{50}, 1 \}.$$