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Question 82 of 105

Q.A random variable X has binomial distribution with n=25n=25 and p=0.8p=0.8, then the standard deviation of X is :

(a) 22
(b) 66
(c) 44
(d) 33
Puducherry TnboardTamil Nadu HSC (DGE) Board 2020MCQ· 1mImportance★★★★★
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For X∼B(25,0.8)X\sim B(25,0.8), the variance npq=4npq=4, so the standard deviation is 4=2\sqrt4=2.

  1. For a binomial random variable XX with parameters nn (number of trials) and pp (probability of success), the variance is Var(X)=npq\mathrm{Var}(X)=npq, where q=1−pq=1-p.
  2. Here n=25n=25 and p=0.8p=0.8, so q=1−0.8=0.2q=1-0.8=0.2.
  3. Compute the variance: Var(X)=npq=25×0.8×0.2\mathrm{Var}(X)=npq=25\times0.8\times0.2. …

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