Question

A r.v. X has the given probability distribution, Find the value of k and calculate mean and variance of X. :
X = x:
-2
-1
0
1
2
3

P (x):
0.1
k
0.2
2k
0.3
k

Answer 1

Answer:

k = 0.1

Mean = 0.8

Step-by-step explanation:

x:    P(x)

-2    0.1

-1    k

0    0.2

1    2k

2    0.3

3    k

0.1 + k + 0.2 + 2k + 0.3 + k =1

=> 4k = 0.4

=> k = 0.1

x:    P(x)

-2    0.1

-1    0.1

0    0.2

1    0.2

2    0.3

3     0.1

Mean = ((-2) * 0.1  + (-1) * 0.1  + 0*0.2 + 1 *0.2 + 2 * 0.3  + 3 * 0.1)

= -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3

= 0.8

Mean = 0.8

Answer 2

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

TO DETERMINE

• The value of k

• The mean and variance for the given Probability distribution of a random variable X

CALCULATION

DETERMINATION OF k

Since a random variable X has the given probability distribution function p(x)

So

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

$\implies \sf{0.1 + k + 0.2 + 2k + 0.3 + k = 1\: }$

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

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

$\implies \sf{ k = 0.1\: }$

$\boxed{\sf{ \: \: Hence \: \: k = 0.1 \: \: \: }}$

CALCULATION OF MEAN

Mean

$= \sf{ \sum xp(x)}$

$= \sf{ ( - 2 \times 0.1) + ( - 1 \times k) + (0 \times 0.2) + (1 \times 2k) + (2 \times 0.3) + (3 \times k)\: }$

$= \sf{ - 0.2 - k + 0 + 2k + 0.6 + 3k\: }$

$= \sf{ 4k + 0.4\: }$

$= \sf{ 0.4 + 0.4\: }$

$= \sf{ 0.8\: }$

CALCULATION OF VARIANCE

Now

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

$= \sf{ ( 4\times 0.1) + ( 1 \times k) + (0 \times 0.2) + (1 \times 2k) + (4 \times 0.3) + (9 \times k)\: }$

$= \sf{ 0.4 + k+ 0 + 2k + 1.2+ 9 k\: }$

$= \sf{ 1.6 + 12 k\: }$

$= \sf{ 1.6 + 1.2 \: }$

$= \sf{ 2.8\: }$

Hence Variance

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

$\sf{ = 2.8 - {(0.8)}^{2} \: }$

$= \sf{ 2.8 - 0.64}$

$= \sf{ 2.16}$

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