Question

The probability distribution of a random variable X is as follows
X=x
1
2
3
4
K
p(x)
0.1
K
0.2
3K
0.3
1) Find the value of k.
2) Find the mean and variance.

Answer 1

Answer:

K = 0.1

Mean = 2.13

Step-by-step explanation:

x      P(x)

1       0.1

2      K

3      0.2

4      3K

K      0.3

Total Probability = 1

=> P(1) + P(2) + P(3) + P(4) + P(K) = 1

01 + K + 0.2 + 3K + 0.3 = 1

=> 4K = 0.4

=> K = 0.1

x      P(x)

1       0.1

2      0.1

3      0.2

4      0.3

0.1    0.3

Mean =  ( 1 * 0.1  + 2 * 0.1  + 3 * 0.2  + 4 * 0.3  + 0.1 * 0.3)

= ( 0.1 + 0.2 + 0.6 + 1.2 +  0.03)

= 2.13

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

For given probability distribution of a random variable X to calculate

1) Find the value of k

2) Find the mean and variance.

CALCULATION

1)

Since p(x) is the probability function of a random variable X

Then

$\sf{ \sum p(x) = 1 \: }$

$\implies \sf{ (0.1 + k + 0.2 + 3k + 0.3) = 1}$

$\implies \sf{ 0.6 + 4k = 1}$

$\implies \sf{ 4k = 0.4}$

$\implies \sf{k = 0.1}$

Hence the value of k is 1

2) CALCULATION OF MEAN

Mean

$= \sf{ \sum \: x \: p(x) \: }$

$= \sf{(1 \times 0.1) + (2 \times k) + (3 \times 0.2) + (4 \times 3k) + (k \times 0.3) \: }$

$\sf{ = 0.1 + 2k + 0.6 + 12k + 0.3k \: }$

$\sf{ = 14.3k + 0.7 \: }$

$\sf{ = (14.3 \times 0.1) + 0.7\: }$

$\sf{ =1.43 + 0.7 \: }$

$\sf{ =2.13 \: }$

Hence Mean = 2.13

CALCULATION OF VARIANCE

Here

$\sf{ \sum \: x \: p(x) = 2.13 \: }$

Again

$\sf{ \sum \: {x}^{2} \: p(x) \: }$

$= \sf{ ( {1}^{2} \times 0.1) + ( {2}^{2} \times k) + ( {3}^{2} \times 0.2) + ( {4}^{2} \times 3k) + ( {k}^{2} + 0.3) \: }$

$= \sf{0.1 + 4k + 1.8 + 48k +( 0.3 \times {k}^{2}) \: }$

$= \sf{1.9 + 52k +( 0.3 \times {k}^{2}) \: }$

$= \sf{1.9 + (52 \times 0.1) + ( 0.3 \times {(0.1)}^{2}) \: }$

$= \sf{1.9 + 5.2 + 0.003 \: }$

$= \sf{7.103 \: }$

Hence variance

$= \sf{ \sum \: {x}^{2} \: p(x) \: - { \bigg( \sum \: {x} \: p(x) \bigg)}^{2} }$

$= \sf{7.103 - {(2.13)}^{2} \: \: }$

$= \sf{ 7.103 - 4.5369\: \: }$

$= \sf{ 2.5661\: \: }$

Hence variance = 2.5661

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