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A First Course in Probability

Sheldon M. Ross

$Math Studyset Vaia Explanations$ Math

9th Edition · 432 pages

ISBN: 9780321794772

Chapter 7: Q.7.31 (page 354)

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

Short Answer

Expert verified

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

Step by step solution

01

Given Information

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

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

02

Explanation

Calculate the variance:

$X = X_{1} + X_{2} + \dots \dots \dots + X_{10}$

$V (X) = V (X_{1}) + V (X_{2}) + \dots \dots \dots + V (X_{10})$

( $∵ X_{i}^{'} s$ are independent )

$= [(E {(X_{1})}^{2} - {(E (X_{1}))}^{2})] + [E {(X_{2})}^{2} - {(E (X_{2}))}^{2}] + \dots \dots \dots + [(E {(X_{10})}^{2} - {(E (X_{10}))}^{2})]$

$= [E {(X_{1})}^{2} + E {(X_{2})}^{2} + \dots \dots \dots + E (X_{10}^{2})] - [{(E (X_{1}))}^{2} + {(E (X_{2}))}^{2} + \dots \dots \dots + {(E (X_{10}))}^{2}]$

03

Explanation

Now $E (X_{i}) = \frac{7}{2} \forall i = 1, 2, \dots \dots, 10$

Calculate the expected value,

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

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

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

$= \frac{91}{6}$

04

Explanation

Calculate the variance of the sum of the rolls:

$∴ V (X) = \frac{10 \times 91}{6} - \frac{10 \times 49}{4}$

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

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

$= \frac{350}{12}$

$= \frac{175}{6}$

05

Final Answer

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

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Most popular questions from this chapter

Show how to compute $Cov (X, Y)$ from the joint moment generating function of $X$ and $Y$ .

Let $U 1, U 2, . . .$ be a sequence of independent uniform $(0, 1)$ random variables. In Example $5 i$ , we showed that for $0 \leq x \leq 1, E [N (x)] = e^{x}$ , where

$N (x) = min \{n : \sum_{i = 1}^{n} U_{i} > x\}$

This problem gives another approach to establishing that result.

(a) Show by induction on n that for $0 < x \leq 1$ 0 and all $n \geq 0$

$P {N (x) \geq n + 1} = \frac{x^{n}}{n!}$

Hint: First condition on $U 1$ and then use the induction hypothesis.

use part (a) to conclude that

$E [N (x)] = e^{x}$

If $X_{1}, X_{2}, \dots, X_{n}$ are independent and identically distributed random variables having uniform distributions over $(0, 1)$ , find

(a) $E [\max (X_{1}, \dots, X_{n})]$ ;

(b) $E [\min (X_{1}, \dots, X_{n})]$ .

In Example 4f, we showed that the covariance of the multinomial random variables $N_{i}$ and $N_{j}$ is equal to $- m P_{i} P_{j}$ by expressing $N_{i}$ and $N_{j}$ as the sum of indicator variables. We could also have obtained that result by using the formula $Var (N_{i} + N_{j}) = Var (N_{i}) + Var (N_{j}) + 2 Cov (N_{i}, N_{j})$

(a) What is the distribution of $N_{i} + N_{j}$ ?

(b) Use the preceding identity to show that $Cov (N_{i}, N_{j}) = - m P_{i}, P_{j}$

Gambles are independent, and each one results in the player being equally likely to win or lose 1 unit. Let W denote the net winnings of a gambler whose strategy is to stop gambling immediately after his first win. Find

(a) P{W > 0}

(b) P{W < 0}

(c) E[W]

See all solutions

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