# 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

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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

The standard deviation of binomial distribution is:

sigma = sqrt(n*p*q)

n - the number of trials

For the given values n=2815,

p - probability of success

q - probability of failure

p=.63, we'll substitute them into the formula:

sigma = sqrt (2815)*(0.63)*(0.63)

Since the probability of failure is not given, we'll put q = p (we know that for a binomial experiment, the probability of failure and success is constant).

sigma = 0.63*sqrt (2815)

sigma = 0.63*53.0565

**sigma = 33.4256**