Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise multiple choice questions question 6 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise multiple choice questions question 6 maths textbook solution

Answers (1)

Answer:

              E\left ( X \right )=\frac{122}{32} or 3.82

Hint:

              To solve this we add all variables

Given:

X 2 3 4 5
P\left ( X \right ) \frac{5}{k} \frac{7}{k} \frac{9}{k}  \frac{11}{k}

Solution:          

Aswe know, \sum_{i=0}^{n} P\left(X_{i}\right)=1

i.e.

\begin{aligned} &P(X=2)+P(X=3)+P(X=4)+P(X=5)=1 \\ &\frac{5}{k}+\frac{7}{k}+\frac{9}{k}+\frac{11}{k}=1 \\ &\frac{5+7+9+11}{k}=1 \\ &k=32 \\ \end{aligned}

E(X)=\sum X \cdot P(X)

X P\left ( X \right ) X.P\left ( X \right )
2 \frac{5}{32} \frac{10}{32}
3 \frac{7}{32} \frac{21}{32}
4 \frac{9}{32} \frac{36}{32}
5 \frac{11}{32} \frac{55}{32}


E(X)=\sum X \cdot P(X)

\begin{aligned} &=\frac{10}{32}+\frac{21}{32}+\frac{36}{32}+\frac{55}{32} \\ &=\frac{122}{32} \end{aligned}

Posted by

Infoexpert

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE, JKBOSE, Assam board exams today; timings, passing marks, key guidelines
anam-khan

CBSE bans use of ChatGPT, mobile during board exam 2025; warns students of action for spreading rumors
anam-khan

Delhi metro to prioritise CBSE Class 10, 12 students for frisking, ticketing during board exam 2025

Latest Question