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 $b = 10$.

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

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

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

In a book of $500$ pages,it is found that there are $250$ typing errors. Assume that Poisson law holds for the number of errors per page. Then,the probability that a random sample of $2$ pages will contain no error,is:

$A$ random variable $X$ has the following probability distribution:
$X$ $1, 2, 3, 4, 5$
$P(X)$ $K^2, 2K, K, 2K, 5K^2$

Then $P(X > 2)$ is equal to:

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

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

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

Find the value of $P(0 < X < 6)$.

If the mean of a Poisson variate $X$ is $1$,then $\sum_{r=0}^{\infty}|r-1| P(X=r)=$

$A$ player tosses two coins. He wins $Rs. 1$ if $1$ head appears,$Rs. 2$ if $2$ heads appear. But he loses $Rs. 3$ if no head appears. The mean of the prize money is

Vedclass Test Series

Exam Paper Generator

Online Exam Module