Math, asked by krishwanshyadav7283, 5 hours ago

Find the value of K, such that the following represents a finite distribution and find its mean and standard deviation x -3 -2 -1 0 1 2 3 p(x) k 2k 3k 4k 3k 2k k

Answers

Answered by phalgunimudlapur85
0

Answer:

Step-by-step explanation:

Solution

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Given that a random variable x has the following probability distribution

X=x  -2  -1  0  1  2 3  

p(x)  0.1  k 0.2   2k  0.3  k

We know that in a probability total probability is unity  

i.e ∑p(x)=1

⇒0.1+k+0.2+2k+0.3+k=1

⇒0.6+4k=1

⇒4k=1−0.6=0.4

4k=0.4

k=  

4

0.4

=0.1

then probability distribution is value of k=0.1

X=x  -2  -1  0  1  2  3

p(x)  0.1  0.1  0.2 0.2  0.3  0.1

we know  

mean of(M) x=  

i=1

n

x  

i

p(x  

i

)

=(−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 of x=0.8

variance of x=∑x  

2

p(x)−μ  

2

 

=(−2)  

2

×0.1+(−1)  

2

×0.1+0  

2

×0.2+1  

2

×0.2+2  

2

×0.3+3  

2

×0.1−(0.8)  

2

 

=0.4+0.1+0+0.2+1.2+0.9−0.64

variance of x=2.16

Answered by abhi185750
0

Answer:

Sum of all probabilities must be equal to 1

So we get K+2K+2K+3K+K

2

+2K

2

+7K

2

+K=1

⇒10K

2

+9K=1

⇒K=−1,

10

1

But probability cannot be negative , therefore the value of K=

10

1

The mean is 1.K+2.2K+3.2K+4.3K+5K

2

+6.2K

2

+7.7K

2

+7K

=30K+66K

2

=3+0.66=3.66

P(0<x<5)=P(x=1)+P(x=2)+P(x=3)+P(x=4)=K+2K+2K+3K=8K=0.8

Step-by-step explanation:

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