Exercise 3.36

Answers

(a)

P (X = \frac{n}{2}) = (\binom{n}{\frac{n}{2}}) {(\frac{1}{2})}^{n}

(b)

Using Sterling’s formula

(\binom{n}{\frac{n}{2}}) = \frac{\sqrt{2 πn} {(\frac{n}{e})}^{n}}{\sqrt{2 π \frac{n}{2}} {(\frac{\frac{n}{2}}{e})}^{\frac{n}{2}} \sqrt{2 π \frac{n}{2}} {(\frac{\frac{n}{2}}{e})}^{\frac{n}{2}}} = \frac{\sqrt{2} 2^{n}}{\sqrt{πn}}

Thus,

P (X = \frac{n}{2}) = \sqrt{\frac{2}{πn}} 2^{n} \frac{1}{2^{n}} = \frac{1}{\sqrt{\frac{πn}{2}}}

fifthist

2021-12-05 00:00