(C) For a probability distribution,the sum of all probabilities must be equal to $1$.Therefore,$0.1 + k + 0.2 + 2k + 0.3 + k = 1$.Combining the terms,

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

(C) For a probability distribution,the sum of all probabilities must be equal to $1$.Therefore,$0.1 + k + 0.2 + 2k + 0.3 + k = 1$.Combining the terms,we get $0.6 + 4k = 1$.$4k = 0.4$,which implies $k = 0.1$.The mean of $X$,denoted by $E(X)$,is given by $\sum x_i P(x_i)$.$E(X) = (-2 \times 0.1) + (-1 \times 0.1) + (0 \times 0.2) + (1 \times 0.2) + (2 \times 0.3) + (3 \times 0.1)$.$E(X) = -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3$.$E(X) = 0.8$.

Class 12 Mathematics Probability Probability distribution

Easy

The following table represents the probability distribution of a random variable $X$ for some $k \in Q$. Find the mean of $X$.
$\begin{array}{|c|c|c|c|c|c|c|} \hline X=x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(X=x) & 0.1 & k & 0.2 & 2k & 0.3 & k \\ \hline \end{array}$

AP EAMCET 2020 All AP EAMCET

A
$1.7$
B
$1.8$
C
$0.8$
D
$0.7$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Probability

Subtopic

Probability distribution

Previous Year

AP EAMCET 2020 Paper All AP EAMCET years →

$A$ random variable $X$ has the range $\{0, 1, 2, \ldots\}$. If $P(X=r) = k(1+r) 3^{-r}$ for $r=0, 1, 2, \ldots$, where $k > 0$ is a real number, then $P(X=0) + P(X=1) + P(X=2) =$

EasyTS EAMCET 2022

View Solution

Let $F(x)$ be the cumulative distribution function (c.d.f.) of a continuous random variable $X$. If $F(b) = 0.7$ and $P(X > a) = 0.4$, then the value of $P(a < X < b)$ is ...

DifficultMHT CET 2026

View Solution

$A$ random variable $X$ takes values $0, 1, 2, 3, \ldots$ with probability $P(X=x) = K(x+1)\left(\frac{1}{5}\right)^x$,where $K$ is a constant. Then $P(X=0)$ is:

EasyAP EAMCET 2021

View Solution

If $X$ is a Poisson variate satisfying the condition $3 P(X=2)=P(X=4)$, then find $P(X=6)$.

DifficultTS EAMCET 2023

View Solution

In a game,$3$ coins are tossed. $A$ person is paid ₹ $7$ if he gets all heads or all tails,and he is supposed to pay ₹ $3$ if he gets one head or two heads. The amount he can expect to win on an average per game is ₹

MediumMHT CET 2023

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

The following table represents the probability distribution of a random variable $X$ for some $k \in Q$. Find the mean of $X$.$\begin{array}{|c|c|c|c|c|c|c|} \hline X=x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(X=x) & 0.1 & k & 0.2 & 2k & 0.3 & k \\ \hline \end{array}$

Similar Questions

$A$ random variable $X$ has the range $\{0, 1, 2, \ldots\}$. If $P(X=r) = k(1+r) 3^{-r}$ for $r=0, 1, 2, \ldots$, where $k > 0$ is a real number, then $P(X=0) + P(X=1) + P(X=2) =$

Let $F(x)$ be the cumulative distribution function (c.d.f.) of a continuous random variable $X$. If $F(b) = 0.7$ and $P(X > a) = 0.4$, then the value of $P(a < X < b)$ is ...

$A$ random variable $X$ takes values $0, 1, 2, 3, \ldots$ with probability $P(X=x) = K(x+1)\left(\frac{1}{5}\right)^x$,where $K$ is a constant. Then $P(X=0)$ is:

If $X$ is a Poisson variate satisfying the condition $3 P(X=2)=P(X=4)$, then find $P(X=6)$.

In a game,$3$ coins are tossed. $A$ person is paid ₹ $7$ if he gets all heads or all tails,and he is supposed to pay ₹ $3$ if he gets one head or two heads. The amount he can expect to win on an average per game is ₹

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The following table represents the probability distribution of a random variable $X$ for some $k \in Q$. Find the mean of $X$.
$\begin{array}{|c|c|c|c|c|c|c|} \hline X=x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(X=x) & 0.1 & k & 0.2 & 2k & 0.3 & k \\ \hline \end{array}$