Chapter 16, Problem 13E

To determine

To find: the standard deviation of the number of children the couple.

Answer to Problem 13E

0.829

Explanation of Solution

Given:

mention that boys and girls are equality. Therefore, probability of having boy or girl is same and is $\frac{1}{2}$

They would not have more than 3 children.

They have children until they get a girl

Formula used:

   $\begin{array}{c}E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}\end{array}$

   $\begin{array}{c}\text{Var}\left(X\right)={\displaystyle \sum {(x-\mu )}^{2}P\left(x\right)}\end{array}$

   $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

Suppose X be number of children. Following is probability model of X

They will have 2 children if first is boy and second is girl

Suppose G shows girl and B shows Boy

  $\begin{array}{c}P\left(BG\right)=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\end{array}$

      $\begin{array}{c}=\frac{1}{4}\end{array}$

They stop with 3 even if 3^rd is boy or girl

  $\begin{array}{c}P\left(x=3\right)=P\left(BBG\right)+P\left(BBB\right)\end{array}$

    $\begin{array}{c}=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\end{array}$

    $\begin{array}{c}=\frac{1}{8}+\frac{1}{8}\end{array}$

    $\begin{array}{c}=\frac{1}{4}\end{array}$

The probability distribution is,

X	1(G)	2(BG)	3(BBG or BBB)
P(X= x)	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{4}$

The expected number of children is

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum xP\left(x\right)}\end{array}$

    $\begin{array}{c}=1\left(\frac{1}{2}\right)+2\left(\frac{1}{4}\right)+3\left(\frac{1}{4}\right)\end{array}$

    $\begin{array}{c}=\frac{1}{2}+\frac{2}{4}+\frac{3}{4}\end{array}$

    $\begin{array}{c}=\frac{7}{4}\end{array}$

    $\begin{array}{c}=1.75\end{array}$

Expected number of children is 1.75

  $\begin{array}{c}\text{Var}\left(X\right)={\displaystyle \sum {\left(x-\mu \right)}^{2}P\left(x\right)}\end{array}$

    $\begin{array}{c}={\left(1-1.75\right)}^2\left(0.5\right)+{\left(2-1.75\right)}^2\left(0.25\right)+{\left(3-1.75\right)}^2\left(0.25\right)\end{array}$

    $\begin{array}{c}=0.5625\left(0.5\right)+0.0625\left(0.25\right)+1.5625\left(0.25\right)\end{array}$

    $\begin{array}{c}=0.6975\end{array}$

The standard deviation is,

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

      $\begin{array}{c}=\sqrt{0.6875}\end{array}$

      $\begin{array}{c}=0.829\end{array}$

Therefore, the standard deviation is 0.829