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The weight of the boys is distributed normally with mean $3.4$ and standard deviation $0.3$, and the weight of the girls is distributed normally with mean $3.2$ and standard deviation $0.3$. If the ratio of the boys and girls is $50:50$, what is the standard deviation of the weight of a baby who is randomly chosen?

My approach is the following:

Let $X$ be the weight of a baby, $B$ be the boys and $G$ be the girls. Then $$X\mid B\sim N(3.4,0.3^2),\ X\mid G\sim N(3.2,0.3^2).$$ What I need to find is $\sigma_X. $

...And at this point I realized that the definitions of $B,G$ are vague. How can I convert this problem into a computable mathematical language?


EDIT:

Let $G$ be a random variable that takes only $2$ values: $M$ for male and $F$ for female. And each has the probability $1\over 2$. That is,

$$P(G=M)={1\over 2},\ P(G=F)={1\over 2}.$$

And if $X$ is the random variable of the weight of a baby, we have

$$(X\mid G=M)\sim N(3.4,0.3^2),\ (X\mid G=F)\sim N(3.2,0.3^2).$$

And what I have to find is $Var(X).$

By the law of total variance,

$$\begin{align}Var(X)&=E(Var(X\mid G))+Var(E(X\mid G))\\ &=E[E(X^2\mid G)-E(X\mid G)^2] + [E(E(X\mid G)^2)-E(E(X\mid G))^2]\\ &=E(E(X^2\mid G))-E(E(X\mid G)^2) + E(E(X\mid G)^2)-E(E(X\mid G))^2\\ &=E(E(X^2\mid G))-E(E(X\mid G))^2.\end{align}$$

And since $E(X\mid G)$ is a random variable depending only on $G$,

$$\begin{align}E(E(X\mid G)) &=P(G=M)\cdot E(X\mid G=M)+P(G=F)\cdot E(X\mid G=F)\\ &= {1\over 2}\cdot 3.4+{1\over 2}\cdot 3.2\\ &= 3.3,\end{align}$$

$$\begin{align}E(E(X^2\mid G)) &=P(G=M)\cdot E(X^2\mid G=M)+P(G=F)\cdot E(X^2\mid G=F) \\ &={1\over 2}\cdot [Var(X\mid G=M)+E(X\mid G=M)^2]+{1\over 2}\cdot [Var(X\mid G=F)+E(X\mid G=F)^2]\\ &={1\over 2}\cdot [0.3^2+3.4^2]+{1\over 2}\cdot [0.3^2+3.2^2]\\ &=10.99 \end{align}$$

Hence

$$Var(X)=10.99-3.3^2=10.99-10.89=0.1$$

And

$$\sigma_X=\sqrt{Var(X)}=0.316227766$$

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    $\begingroup$ this is like a mixture model $f(x)=.5 N(3.4,.09)+.5N(3.2,.09)$. is it? $\endgroup$
    – Masoud
    Mar 27 '19 at 1:42
  • $\begingroup$ @masoud Right, then is it correct to say $f(x)\sim N(\frac{3.4+3.2}{2},\frac{0.09+0.09}{2})=N(3.3,0.09)$? $\endgroup$
    – user642721
    Mar 27 '19 at 1:55
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    $\begingroup$ To calculate sd you could use the law of total variance. $\endgroup$
    – Masoud
    Mar 27 '19 at 2:29
  • $\begingroup$ In general your model be like $p n(a,b) +qn(c,d)$(check identifability) and estimate all parameters include p. even p can be random variable. that in your case the values be .5 . $\endgroup$
    – Masoud
    Mar 27 '19 at 2:35
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    $\begingroup$ You could but you should not. (Could means you could have a bad analysis) It is a bad option. .This ia a mixture model. The target distribution is mixes of two distributions. For example it has two mods and difference from a unimodel normal. you lose the information that may give u more better estimation. If you do that, you assumes that every observation have a same mean and variance that is not correct(and same distribution). $\endgroup$
    – Masoud
    Mar 27 '19 at 2:47
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Let $(X\mid G=m)\sim\mathcal N(\mu_m,\sigma_m^2)$ and $(X\mid G=f)\sim\mathcal N(\mu_f,\sigma_f^2)$ and $G\sim\mathcal U\{m,f\}$

So to verify your calculations:

$$\begin{align}\mathsf E(X)&=\mathsf E\mathsf E (X\mid G)\\&=\tfrac 12(\mu_m+\mu_f)\\&=\tfrac 12(3.4+3.2)\\&=3.3\\[3ex]\mathsf {Var}(X)&=\mathsf{Var}\mathsf E(X\mid G)+\mathsf E\mathsf{Var}(X\mid G)\\&= \mathsf E\mathsf E^2(X\mid G)-\mathsf E^2\mathsf E(X\mid G)+\mathsf E\mathsf{Var}(X\mid G)\\&=\tfrac 12(\mu_m^2+\mu_f^2)-(\tfrac 12(\mu_m+\mu_f))^2+\tfrac 12(\sigma_m^2+\sigma_f^2)\\&=\tfrac 12(3.4^2+3.2^2)-(\tfrac 12(3.4+3.2))^2+0.3^2\\&= 10.90-10.89+0.09\\&=0.10\end{align}$$

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