Q.7.31 - A First Course in Probability

Question

In Problem 7.6, calculate the variance of the sum of the rolls.

Step-by-Step Solution

Answer

The variance of the sum of the rolls are $=\frac{175}{6}$.

Step 1: Given Information

The number of times fair die rolled $=10$times

The variance of the sum of the rolls$=?$

Step 2: Explanation

Calculate the variance:

$X=X_1+X_2+\dots \dots \dots +X_{10}$

$V\left(X\right)=V\left(X_1\right)+V\left(X_2\right)+\dots \dots \dots +V\left(X_{10}\right)$

 ($\because X_i^{\text{'}}s$ are independent )

$=\left[\left(\mathrm{E}{\left(X_1\right)}^2-{\left(E\left(X_1\right)\right)}^2\right)\right]+\left[E{\left(X_2\right)}^2-{\left(E\left(X_2\right)\right)}^2\right]+\dots \dots \dots +\left[\left(\mathrm{E}{\left(X_{10}\right)}^2-{\left(E\left(X_{10}\right)\right)}^2\right)\right]$

$=\left[E{\left(X_1\right)}^2+E{\left(X_2\right)}^2+\dots \dots \dots +E\left(X_{10}^2\right)\right]-\left[{\left(\mathrm{E}\left(X_1\right)\right)}^2+{\left(E\left(X_2\right)\right)}^2+\dots \dots \dots +{\left(E\left(X_{10}\right)\right)}^2\right]$

Step 3: Explanation

Now $\mathrm{E}\left({\mathrm{X}}_{\mathrm{i}}\right)=\frac{7}{2} \forall \mathrm{i}=1,2,\dots \dots ,10$

Calculate the expected value,

$E\left(X_i^2\right)=\frac{1}{6}\left[1^2+2^2+\dots \dots ..+6^2\right]\forall i=1,2,\dots \dots ,10$

$=\frac{1}{6}\frac{6(6+1)(12+1)}{6}$

$=\frac{1}{6}\times 13\times 7$

$=\frac{91}{6}$

Step 4: Explanation

Calculate the variance of the sum of the rolls:

$\therefore \mathrm{V}\left(\mathrm{X}\right)=\frac{10\times 91}{6}-\frac{10\times 49}{4}$

$=\frac{20\times 91-30\times 49}{12}$

$=\frac{1820-1470}{12}$

$=\frac{350}{12}$

$=\frac{175}{6}$

Step 5: Final Answer

Therefore, the variance of the sum of the rolls is $=\frac{175}{6}$