Aspire Faculty ID #17246 · Topic: CUET UG 2022 Applied Mathematics · Just now
CUET UG 2022 Applied Mathematics

The probability distribution of $X$ is:
 x 01 2 3 4 
 p(X= x) 0.12k k k 2k 
 then var(X) = 

Solution

Since total probability = 1

$0.1+2k+k+k+2k=1$

$0.1+6k=1$

$6k=0.9$

$k=0.15$

Now probabilities:

$P(0)=0.1$

$P(1)=0.3$

$P(2)=0.15$

$P(3)=0.15$

$P(4)=0.3$

Now,

$E(X)=0(0.1)+1(0.3)+2(0.15)+3(0.15)+4(0.3)$

$=0+0.3+0.3+0.45+1.2$

$=2.25$

Now,

$E(X^2)=0+1(0.3)+4(0.15)+9(0.15)+16(0.3)$

$=0.3+0.6+1.35+4.8$

$=7.05$

Variance:

$\mathrm{Var}(X)=E(X^2)-[E(X)]^2$

$=7.05-(2.25)^2$

$=7.05-5.0625$

$=1.9875$

$= \frac{159}{80}$
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