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Find the standard deviation, for the binomial distribution which has the stated values of n and p.
Round your answer to the nearest hundredth.
Problem - n=2815, p=.63
Solution
We know for a binomial distibution with probability function,
P(x = r) = nCr * p^r *(1-p)^(n-r), the mean = np and variance = np(1-p).
Standard deviation, s= (variance)^(1/2) ={np(1-p)}^(1/2).
Standard deviation , s = (np(1-p)}^(1/2)....(1)
So we substitute n = 2815, p = 0.63, in the formula at (1).
Standard deviation = {2815*0.63*(1-0.63)}^(1/2).
Standard deviation = {2815*0.63*0.37}^(1/2) = 25.6159 = 25.62.
So the required standard deviation for the give binomial distribution is 25.62

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Find the standard deviation, for the binomial distribution which has.pdf

  • 1. Find the standard deviation, for the binomial distribution which has the stated values of n and p. Round your answer to the nearest hundredth. Problem - n=2815, p=.63 Solution We know for a binomial distibution with probability function, P(x = r) = nCr * p^r *(1-p)^(n-r), the mean = np and variance = np(1-p). Standard deviation, s= (variance)^(1/2) ={np(1-p)}^(1/2). Standard deviation , s = (np(1-p)}^(1/2)....(1) So we substitute n = 2815, p = 0.63, in the formula at (1). Standard deviation = {2815*0.63*(1-0.63)}^(1/2). Standard deviation = {2815*0.63*0.37}^(1/2) = 25.6159 = 25.62. So the required standard deviation for the give binomial distribution is 25.62