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
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
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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