For a binomial distribution with mean 4 and standard deviation \(\sqrt{3}\), what is the value of \(P(X=0)\)?
\(\left(\dfrac{3}{4}\right)^{16}\)
For a binomial distribution, mean \(np=4\) and variance \(npq=3\), so \(q=\frac{npq}{np}=\frac{3}{4}\) and \(p=\frac{1}{4}\), giving \(n=\frac{np}{p}=16\). Then \(P(X=0)={}^{n}C_{0}\,p^{0}q^{n}=q^{n}=\left(\frac{3}{4}\right)^{16}\).
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If log e(x) is normally distributed with mean 1 and variance 4, then P(0.5 < x < 2) is:
(where the area between z = 0 and z = 0.25 is 0.0987 and the area between z = 0 and z = 0.5 is 0.1915)
