Question

in a binomial distribution with 5 trials mean is 3 , find parameter

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Answer 1

Step-by-step explanation:

Let n and p be the parameters of the distribution. Then,

Mean = np and Variance = npq

Given,

n=5 and, Mean + Variance = 1.8

np+npq=1.8

5p+5pq=1.8

5p+5p(1−p)=1.8

5p

2

−10p+1.8=0

p

2

−2p+0.36=0

(p−0.2)(p−1.8)=0

p=0.2

Thus,

n=5,p=0.2,q=0.8

Thus, if X denotes the binomial variate, then

P(X=r)=5

C

r

(0.2)

r

(0.8)

5−r

,r=0,1,2,3,4,5.

This is the required binomial distribution.

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