If a random variable X has p.d.f. f(x) = begi | vedclass

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

Question

(B) For a probability density function (p.d.f.),the total area under the curve must be $1$. Thus,$\int_{1}^{3} f(x) dx = 1$. Substituting $f(x) = \fra

Vedclass · Accepted Answer

$-9, 10$

Vedclass · Answer

(B) For a probability density function (p.d.f.),the total area under the curve must be $1$. Thus,$\int_{1}^{3} f(x) dx = 1$. Substituting $f(x) = \frac{ax^2}{2} + bx$: $\int_{1}^{3} (\frac{ax^2}{2} + bx) dx = [\frac{ax^3}{6} + \frac{bx^2}{2}]_{1}^{3} = 1$. Evaluating at the limits: $(\frac{27a}{6} + \frac{9b}{2}) - (\frac{a}{6} + \frac{b}{2}) = 1$. $\frac{26a}{6} + \frac{8b}{2} = 1 \implies \frac{13a}{3} + 4b = 1 \implies 13a + 12b = 3$ (Equation $1$). Given $f(2) = 2$: $\frac{a(2)^2}{2} + b(2) = 2 \implies 2a + 2b = 2 \implies a + b = 1 \implies b = 1 - a$ (Equation $2$). Substitute Equation $2$ into Equation $1$: $13a + 12(1 - a) = 3$. $13a + 12 - 12a = 3$. $a = 3 - 12 = -9$. Using $b = 1 - a$: $b = 1 - (-9) = 10$. Thus,$a = -9$ and $b = 10$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

If a random variable $X$ has p.d.f. $f(x) = \begin{cases} \frac{ax^2}{2} + bx & , \text{if } 1 \leqslant x \leqslant 3 \\ 0 & , \text{otherwise} \end{cases}$ and $f(2) = 2$,then the values of $a$ and $b$ are,respectively

Similar Questions

The distribution of a random variable $X$ is given below. The value of $k$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X = x)$ $\frac{1}{10}$ $k$ $\frac{1}{5}$ $2k$ $\frac{3}{10}$ $k$

In a city,$10$ accidents take place in a span of $50$ days. Assuming that the number of accidents follows the Poisson distribution,the probability that three or more accidents occur in a day is:

If the probability mass function (p.m.f.) is given by $P(X) = k \binom{4}{x}$ for $x = 0, 1, 2, 3, 4$ and $k > 0$,and $P(X) = 0$ otherwise,then the value of $k$ is:

In a game,a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decides to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins or loses. (in $/216$)

$A$ six-faced fair die is thrown until $2$ appears. What is the probability that $2$ appears in an even number of trials? (The die has six faces numbered $1, 2, 3, 4, 5,$ and $6$.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module