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Mean and variance of a random variable — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMean and variance of a random variable4 Marks
Question
Two cards are selected at random from a box which contains five cards numbered 1, 1, 2, 2, and 3. Let X denote the sum and Y the maximum of the two numbers drawn. Find the probability distribution, mean and variance of X and Y.
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Answer

Box contains five cards 1, 1, 2, 2, 3.

Here,

X denotes the sum of the two number on cards drawn.

Y denotes the maximum of the two number.

So, X = 2, 3, 4, 5

Y = 1, 2, 3

P(X = 2) = P(1)P(1)

$=\frac{2}{5}\times\frac{1}{4}$

$=0.1$

P(X = 3) = P(1)P(2) + P(2)P(1)

$=\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}$

$=0.4$

P(X = 4) = P(2)P(2) + P(1)P(3) + P(3)P(1)

$=\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}$

$=0.3$

P(X = 5) = P(2)P(3) + P(3)P(2)

$=\frac{2}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$

$=0.2$

Probability distribution for X

x: 2 3 4 5
P(x): 0.1 0.4 0.3 0.2
Now,

xi pi xipi xi2pi
2 0.1 0.1 0.4
3 0.4 1.2 3.6
4 0.3 1.2 4.8
5 0.2 1.0 5.0
    $\sum\text{xp}=3.6$

 $\sum\text{x}^2\text{p}=13.8$
Mean $=\sum\text{xp}$

Mean = 3.6

Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2$

$=13.8-(3.6)^2$

$=13.8-12.96$

Variance = 0.84

P(Y = 1) = P(1)P(1)

$=\frac{2}{5}\times\frac{1}{4}$

$=\frac{2}{20}$

$=0.1$

P(Y = 2) = P(1)P(2) + P(2)P(1) + P(2)P(2)

$=\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{1}{4}$

$=0.5$

P(Y = 3) = P(1)P(3) + P(2)P(3) + P(3)P(1) + P(3)P(2)

$=\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$

$=0.4$

Probability distribution for Y is

x:
1
2
3
p(x):
0.1
0.5
0.4
 
yi pi yipi yi2pi
1 0.1 0.1 0.1
2 0.5 1.0 2.0
3 0.4 1.2 3.6
 

 

 $\sum\text{xp}=2.3$

 $\sum\text{x}^2\text{p}=5.7$
Mean $=\sum\text{xp}=2.3$

Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2$

$=5.1-(2.3)^2$

Variance = 0.41

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