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What are the mean and standard deviation of a binomial probability distribution with #n=120 # and #p=3/4 #?

Answer 1

Ezekiel Anderson

Mean #90#
Standard deviation #sqrt{45/2}#

#mu = np#

#= (120)*(3/4)#

#= 90#

#sigma = sqrt{np(1-p)}#

#= sqrt{120*(3/4)*(1/4)}#

#= 3sqrt{5/2}#

If the random variable #X# follows the binomial distribution with parameters #n in NN# and #p in [0,1]#, the probability of getting exactly #k# successes in #n# trials, is given by

#"Pr"(X=k) = nC_k*p^k*(1-p)^{n-k}#

The mean is calculated from #"E"(X)#.

The standard deviation comes from the square root of the variance, #"Var"(X) = "E"(X^2)-["E"(X)]^2#

Where

#"E"(X) = sum_{k=0}^n k*"Pr"(X=k)#

#"E"(X^2) = sum_{k=0}^n k^2*"Pr"(X=k)#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.