Question 1157914: . For a Binomial Distribution, mean is 6 and standard deviation is √2 . Find the parameters
Answer by jim_thompson5910(35256)   (Show Source):
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Answers:
n = 9
p = 0.67 approximately (exact value is p = 2/3)
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Work Shown:
mu = mean
mu = n*p
mu = 6
n*p = 6
p = 6/n
sigma = standard deviation
sigma = sqrt(n*p*(1-p))
sqrt(2) = sqrt(n*p*(1-p))
2 = n*p*(1-p) ... square both sides
2 = 6*(1-p) ... plug in n*p = 6
2 = 6*(1-6/n) ... plug in p = 6/n
2 = 6-36/n .... distribute
2n = 6n-36 .... multiply every term by n to clear out the denominator
2n-6n = -36 ... subtract 6n from both sides
-4n = -36
n = -36/(-4) ... divide both sides by -4
n = 9
With that value of n, we can find the corresponding value of p
p = 6/n
p = 6/9
p = 2/3
p = 0.67 approximately
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As a check,
mu = n*p
mu = 9*(2/3)
mu = 18/3
mu = 6
If you use the approximate value of p, then mu will be approximate as well
mu = n*p
mu = 9*0.67
mu = 6.03
The same applies to the standard deviation
If you use n = 9 and p = 2/3, then,
sigma = sqrt(n*p*(1-p))
sigma = sqrt(9*(2/3)*(1-2/3))
sigma = sqrt(2)
sigma = 1.41421356 which is approximate
or if you use p = 0.67 instead, then,
sigma = sqrt(n*p*(1-p))
sigma = sqrt(9*0.67*(1-0.67))
sigma = sqrt(1.9899)
sigma = 1.41063815 also approximate; fairly close to the previous sigma value computed above
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