CONTEXT: First year university statistics course exam question

Suppose couples decide to have children until they either have a child of each sex, or they have three children. Assume that births are independent and equally likely to be boys or girls. I calculated the expected number of girls per couple, $E(X)$, to be $1.25$.

A gathering is organised where $50$ couples who followed this approach bring their children along. What is the approximate distribution for modelling the total number of girls at the gathering?

(a) Binomial(50, 0.5)

(b) Binomial(50, 0.625)

(c) Normal(62.5, 5.86)

(d) Normal(62.5, 41.45)

(e) Uniform(0, 150)

I know the correct answer is (c), but don't understand why it is a normal distribution instead of a binomial. Shouldn't it be normal since the amount of girls you can have is discrete (as you can't have 0.5 of a person)? But then I also understand that $n\cdot E(X) = 62.5 = \mu $ which is not discrete. Is the fact that this isn't a discrete number what makes the distribution normal?

Any guidance would be greatly appreciated :)

  • $\begingroup$ This question requires the self study tag. $\endgroup$ Commented Sep 9, 2019 at 13:11
  • 1
    $\begingroup$ This exercise asks for an approximation. Yes, you could approximate the distribution with a Binomial--but that would still be an approximation. (The exact answer is a little complicated, which is the implicit motivation for seeking an approximation.) $\endgroup$
    – whuber
    Commented Sep 9, 2019 at 13:39
  • $\begingroup$ Might I ask how you calculated $\mathbb{E}[X]$ to be $1.25$ ? $\endgroup$
    – Emil
    Commented Sep 9, 2019 at 14:09
  • $\begingroup$ @Emil Due to the assumption, I found probability of a girl is 0.5. So the probability of three girls is P(GGG) = (1/2)^3 = 0.125. So, P(X=3) where X is the a girl is 0.125. For P(X=1) we have P(GB, BG, BBG) = (0.5)^2 + (0.5)^2 + (0.5)^3 = 0.625. I did this for P(X = 0, 1, 2, 3) and that gave me E(X) = 0(0.125)+1(0.625)+2(0.125)+3(0.125) = 1.25. $\endgroup$
    – Ruby Pa
    Commented Sep 10, 2019 at 3:13
  • $\begingroup$ @whuber So what would an approximate binomial distribution look like for this question? Is the process to arrive at this a lot more complicated than it would be for a normal, and that's a possible indication as to why normal is used? $\endgroup$
    – Ruby Pa
    Commented Sep 10, 2019 at 3:15

1 Answer 1


Part 1

Unless I am grossly misinterpreting or missing something very obvious (in which case I obviously welcome any correction), I wouldn't agree that the correct answer is (c). In fact, I wouldn't even agree that the question is well defined. In my opinion, if the question was presented exactly like this at the exam and without further info, then it is unsuitable and ill-posed. For an explanation of that, please see Part 2.

For starters, using the probabilities shown further below, we calculate that $$\mu = \mathbb{E}[X] = \sum_x x\mathbb{P}[X=x] = 1.25,$$ $$\sigma^2 = \mathbb{V}[X] = \left(\sum_{x}x^2\mathbb{P}[X=x]\right) - \mu^2 = 0.6875$$ Therefore, if $Y = 50X$, then $\mu_Y = \mathbb{E}[50X] = 50\mu = 50\cdot 1.25 = 62.5$, while $\sigma_Y^2 = \mathbb{V}[50X] = 50^2\sigma^2 = 50^2\cdot 0.6875 = 1,718.75$.

You have successfully calculated that $\mu_Y = 62.5$, so all that's left is "matching" it to one of the possible values of (c) and (d). As $\sqrt{1,718.15}\approx 41.45$, you can see that the most likely answer is (d), assuming that the question specifies a normal distribution in terms of mean and standard deviation, i.e. $\mathcal{N}(\mu,\sigma)$ instead of mean and variance, i.e. $\mathcal{N}(\mu, \sigma^2)$ (this should have been mentioned in the lecture notes or the description of the question on the exam).

Your confusion about why the final distribution is not binomial but normal is a bit misplaced. The final distribution is binomial, however the binomial distribution is impractical to calculate when $n$ is large. As it happens, in those cases of a large $n$, the binomial distribution with parameters $(n, p)$ can be approximated by a normal distribution with mean $np$ and variance $np(1-p)$, and the larger the $n$ the better the approximation. For some more details on this, check out the Wiki page.

Part 2

The reason I said the most likely (but not correct) answer is (d) because the distribution of the amount of girls, $X$ is not binomially distributed, because you don't know beforehand what the $n$ parameter will be, i.e. after how many trials the experiment will end. Let's look at what the sample space of three children looks like, depending on what the first child is:

$$\newcommand\pad[1]{\hspace{-18pt}\llap{#1}\phantom{1764}} \hspace{18pt}\begin{matrix} &&&&&&\pad{B}\\ &&&&&\pad{B}&\pad{}&\pad{G}\\ &&&&\pad{B}&\pad{}&\pad{G}&\pad{}&\pad{G} \end{matrix} \hspace{18pt}\begin{matrix} &&&&&&\pad{G}\\ &&&&&\pad{G}&\pad{}&\pad{B}\\ &&&&\pad{G}&\pad{}&\pad{B}&\pad{}&\pad{G} \end{matrix} $$ As it can be seen from here, we can calculate the probabilities of $\{X=x\}$ as follows:

$\mathbb{P}[X = 0] = \mathbb{P}[\text{3 kids, all boys}] = 0.125, $

$\mathbb{P}[X = 1] = \mathbb{P}[\text{2 kids, 1 girl}] + \mathbb{P}[\text{3 kids, 1 girl}]= 0.625, $

$\mathbb{P}[X = 2] = \mathbb{P}[\text{3 kids, 2 girls, in the order BBG}] = 0.125, $

$\mathbb{P}[X = 3] = \mathbb{P}[\text{3 kids, all girls}] = 0.125 $

You calculated these probabilities yourself, of course, as you showed in the comment, but the reason I wrote the calculations nonetheless is because as I mentioned above, we do not know beforehand the number of trials it will take to achieve the desired result, but there is also an upper bound of the amount of trials you can have (in this case, three trials). I would therefore classify this as some kind of bounded negative binomial distribution problem, with an example of the unbounded case seen here.


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