Given the probability density function: f(x) | vedclass

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

Question

(A) Given the p.d.f. of a random variable $X$: $f(x) = \begin{cases} 3(1 - 2x^2), & 0 < x < 1 \ 0, & 	ext{otherwise} \end{cases}$ The probability is

Vedclass · Accepted Answer

$\frac{179}{864}$

Vedclass · Answer

(A) Given the p.d.f. of a random variable $X$: $f(x) = \begin{cases} 3(1 - 2x^2), & 0 The probability is calculated as: $P\left(\frac{1}{4} $= 3 \left[ x - \frac{2}{3}x^3 ight]_{1/4}^{1/3}$ $= 3 \left[ \left( \frac{1}{3} - \frac{2}{3} \cdot \frac{1}{27} ight) - \left( \frac{1}{4} - \frac{2}{3} \cdot \frac{1}{64} ight) ight]$ $= 3 \left[ \left( \frac{1}{3} - \frac{2}{81} ight) - \left( \frac{1}{4} - \frac{1}{96} ight) ight]$ $= 3 \left[ \frac{27-2}{81} - \frac{24-1}{96} ight] = 3 \left[ \frac{25}{81} - \frac{23}{96} ight]$ $= 3 \left[ \frac{25 	imes 32 - 23 	imes 27}{2592} ight] = 3 \left[ \frac{800 - 621}{2592} ight]$ $= 3 	imes \frac{179}{2592} = \frac{179}{864}$

$X = x$	$0$	$1$	$2$
$P(X = x)$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{2}{5}$

$X = x$	$0$	$1$	$2$	$3$	$4$
$P(X = x)$	$k$	$2k$	$4k$	$2k$	$k$

Given the probability density function: $f(x) = \begin{cases} 3(1 - 2x^2), & 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$ The probability $P\left(\frac{1}{4} < X < \frac{1}{3}\right)$ is given by: $P\left(\frac{1}{4} < X < \frac{1}{3}\right) = \int_{1/4}^{1/3} 3(1 - 2x^2) \, dx$

Similar Questions

The probability distribution of a random variable $X$ is given by
$X = x$ $0$ $1$ $2$
$P(X = x)$ $\frac{1}{5}$ $\frac{2}{5}$ $\frac{2}{5}$

Then the variance of $X$ is

The probability distribution of a random variable $X$ is given by

$X=x_i$ $0$ $1$ $2$ $3$ $4$

$P(X=x_i)$ $0.4$ $0.3$ $0.1$ $0.1$ $0.1$

Then the variance of $X$ is

In a box,there are $20$ cards,out of which $10$ are labelled as $A$ and the remaining $10$ are labelled as $B$. Cards are drawn at random,one after the other and with replacement,until a second $A$-card is obtained. The probability that the second $A$-card appears before the third $B$-card is

An observer counts $240$ vehicles per hour at a specific location on a highway. Assuming that the arrival of vehicles at the location follows a Poisson distribution, the probability that more than two vehicles arrive over a $30 \text{ sec}$ time interval is

The probability distribution of a discrete random variable $X$ is given by the following table:
$X = x$ $0$ $1$ $2$ $3$ $4$
$P(X = x)$ $k$ $2k$ $4k$ $2k$ $k$

Then the value of $P(X \leq 2)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module