(A) For a discrete random variable,the sum of probabilities must be $1$. $\sum_{x=0}^{6} f(x) = 1$ $k(0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = 1$ $k

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

(A) For a discrete random variable,the sum of probabilities must be $1$. $\sum_{x=0}^{6} f(x) = 1$ $k(0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = 1$ $k(0 + 1 + 4 + 9 + 16 + 25 + 36) = 1$ $91k = 1 \implies k = \frac{1}{91}$ The cumulative distribution function $F(4)$ is defined as $P(X \leq 4)$. $F(4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)$ $F(4) = k(0^2 + 1^2 + 2^2 + 3^2 + 4^2)$ $F(4) = k(0 + 1 + 4 + 9 + 16) = 30k$ Substituting $k = \frac{1}{91}$,we get: $F(4) = 30 \times \frac{1}{91} = \frac{30}{91}$

Class 12 Mathematics Probability Probability distribution

Medium

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:

MHT CET 2023 All MHT CET

A
$\frac{30}{91}$
B
$\frac{30}{97}$
C
$\frac{15}{47}$
D
$\frac{15}{97}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Probability

Subtopic

Probability distribution

Previous Year

MHT CET 2023 Paper All MHT CET years →

$A$ random variable $X$ has the following probability distribution:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $k-1$ $3k$ $k$ $3k$ $3k^2$ $k^2$ $k^2+k$

Then the value of $k$ is:

MediumKCET 2019

View Solution

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

MediumAP EAMCET 2008

View Solution

The probability distribution of a discrete random variable $X$ is given by the following table:
$X$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X)$ $K$ $2K$ $3K$ $4K$ $5K$ $6K$

Find the value of $P(2 < X < 6)$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X)$	$K$	$2K$	$3K$	$4K$	$5K$	$6K$

MediumMHT CET 2021

View Solution

If the probability distribution of a random variable $X$ is as follows,then the mean of $X$ is:
$X = x_i$ $-1$ $0$ $1$ $2$
$P(X = x_i)$ $k^3$ $2k^3 + k$ $4k - 10k^2$ $4k - 1$

$X = x_i$	$-1$	$0$	$1$	$2$
$P(X = x_i)$	$k^3$	$2k^3 + k$	$4k - 10k^2$	$4k - 1$

MediumAP EAMCET 2025

View Solution

Which of the following can not be a valid assignment of probabilities for outcomes of sample space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$?
Outcome Probability
$\omega_{1}$ $0.1$
$\omega_{2}$ $0.01$
$\omega_{3}$ $0.05$
$\omega_{4}$ $0.03$
$\omega_{5}$ $0.01$
$\omega_{6}$ $0.2$
$\omega_{7}$ $0.6$

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

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

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

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.01$
$\omega_{3}$	$0.05$
$\omega_{4}$	$0.03$
$\omega_{5}$	$0.01$
$\omega_{6}$	$0.2$
$\omega_{7}$	$0.6$

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:

Similar Questions

$A$ random variable $X$ has the following probability distribution:$X$$1$$2$$3$$4$$5$$6$$7$$P(X)$$k-1$$3k$$k$$3k$$3k^2$$k^2$$k^2+k$Then the value of $k$ is:

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$

The probability distribution of a discrete random variable $X$ is given by the following table:$X$$1$$2$$3$$4$$5$$6$$P(X)$$K$$2K$$3K$$4K$$5K$$6K$Find the value of $P(2 < X < 6)$.

If the probability distribution of a random variable $X$ is as follows,then the mean of $X$ is: $X = x_i$$-1$$0$$1$$2$$P(X = x_i)$$k^3$$2k^3 + k$$4k - 10k^2$$4k - 1$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ random variable $X$ has the following probability distribution:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $k-1$ $3k$ $k$ $3k$ $3k^2$ $k^2$ $k^2+k$

Then the value of $k$ is:

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

The probability distribution of a discrete random variable $X$ is given by the following table:
$X$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X)$ $K$ $2K$ $3K$ $4K$ $5K$ $6K$

Find the value of $P(2 < X < 6)$.

If the probability distribution of a random variable $X$ is as follows,then the mean of $X$ is:
$X = x_i$ $-1$ $0$ $1$ $2$
$P(X = x_i)$ $k^3$ $2k^3 + k$ $4k - 10k^2$ $4k - 1$