Let X denote the sum of the numbers obtained | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) When two fair dice are rolled,$6 	imes 6 = 36$ outcomes are possible.The probability distribution of the sum $X$ is:$X$$2$$3$$4$$5$$6$$7$$8$$9$$1

Vedclass · Accepted Answer

Variance $= 5.833$,Standard Deviation $= 2.415$

Vedclass · Answer

(A) When two fair dice are rolled,$6 	imes 6 = 36$ outcomes are possible.The probability distribution of the sum $X$ is:$X$$2$$3$$4$$5$$6$$7$$8$$9$$10$$11$$12$$P(X)$$\frac{1}{36}$$\frac{2}{36}$$\frac{3}{36}$$\frac{4}{36}$$\frac{5}{36}$$\frac{6}{36}$$\frac{5}{36}$$\frac{4}{36}$$\frac{3}{36}$$\frac{2}{36}$$\frac{1}{36}$$E(X) = \sum X_i P(X_i) = 2(\frac{1}{36}) + 3(\frac{2}{36}) + 4(\frac{3}{36}) + 5(\frac{4}{36}) + 6(\frac{5}{36}) + 7(\frac{6}{36}) + 8(\frac{5}{36}) + 9(\frac{4}{36}) + 10(\frac{3}{36}) + 11(\frac{2}{36}) + 12(\frac{1}{36}) = 7$.$E(X^2) = \sum X_i^2 P(X_i) = 4(\frac{1}{36}) + 9(\frac{2}{36}) + 16(\frac{3}{36}) + 25(\frac{4}{36}) + 36(\frac{5}{36}) + 49(\frac{6}{36}) + 64(\frac{5}{36}) + 81(\frac{4}{36}) + 100(\frac{3}{36}) + 121(\frac{2}{36}) + 144(\frac{1}{36}) = \frac{1974}{36} = \frac{329}{6} \approx 54.833$.$	ext{Var}(X) = E(X^2) - [E(X)]^2 = 54.833 - 49 = 5.833$.$	ext{Standard Deviation} = \sqrt{	ext{Var}(X)} = \sqrt{5.833} \approx 2.415$.

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$3k$	$3k^2$	$k^2$	$2k^2$	$7k^2+k$

Let $X$ denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of $X$.

Similar Questions

$A$ random variable $X$ has the following probability distribution:

$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X)$ $0$ $k$ $2k$ $3k$ $3k^2$ $k^2$ $2k^2$ $7k^2+k$

Determine $P(X > 6)$.

If the probability function of a random variable $X$ is given by $P(X=n) = \frac{k(n+1)}{3^n}$ for $n \in \mathbb{N} \cup \{0\}$ where $k$ is a constant, then $P(X < 2) = $

$15$ coupons are numbered from $1$ to $15$. Seven coupons are selected at random with replacement. What is the probability that the maximum number on the selected coupons is $9$?

In a game,a man wins $₹ 40$ if he gets $5$ or $6$ on a throw of a fair die and loses $₹ 20$ for getting any other number on the die. If he decides to throw the die either until he gets a $5$ or $6$ or to a maximum of $3$ throws,then his expected gain/loss (in rupees) is:

The p.d.f. of a discrete random variable $X$ is defined as $f(x) = \begin{cases} kx^2, & x \in \{0, 1, 2, 3, 4, 5, 6\} \\ 0, & \text{otherwise} \end{cases}$. Then the value of $F(4)$ (c.d.f.) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module