Math, asked by Shubhasmita31, 2 months ago

Find the so mean and variance of the uniform for probability distribution given by f(x)=1/n for x = 1, 2, 3 ...n​

Answers

Answered by poonammishra148218
0

Answer:

The variance of uniform random variable X is,

V(X)=\frac{(n-1)(n+1)}{12}$$

Step-by-step explanation:

Step 1: The given uniform probability distribution is,

$$f(x)=1 / n f o r x=1,2,3, \ldots, n$$

The mean of uniform random variable X is,

$$\begin{aligned}E(X) & =\sum_{x=1}^n x \cdot f(x) \\& =\sum_{x=1}^n x \cdot\left(\frac{1}{n}\right) \\& =\frac{1}{n} \sum_{x=1}^n x \quad \quad\left(\text { Since } \sum_{x=1}^n x=\frac{n(n+1)}{2}\right) \\& =\frac{1}{n}\left(\frac{n(n+1)}{2}\right) \quad \quad \quad(\quad) \\& =\frac{n+1}{2}\end{aligned}$$

Therefore, the mean of uniform random variable X is,

E(X)=\frac{n+1}{2}$$

Step 2:The variance of uniform random variable X is,

$$\begin{aligned}V(X) & =E\left(X^2\right)-[E(X)]^2 \\& =\sum_{x=1}^n x^2 \cdot f(x)-\left[\frac{n+1}{2}\right]^2 \\& =\sum_{x=1}^n x^2 \cdot\left(\frac{1}{n}\right)-\left[\frac{n+1}{2}\right]^2 \\& =\frac{1}{n} \sum_{x=1}^n x^2-\left[\frac{n+1}{2}\right]^2\end{aligned}$$$\left(\right.$ Since $\left.\sum_{x=1}^n x^2=\frac{n(n+1)(2 n+1)}{6}\right)$

\begin{aligned}& =\frac{1}{n}\left(\frac{n(n+1)(2 n+1)}{6}\right)-\left[\frac{n+1}{2}\right]^2 \\& =\left(\frac{(n+1)(2 n+1)}{6}\right)-\left(\frac{(n+1)^2}{4}\right) \\& =\left(\frac{2 n^2+n+2 n+1}{6}\right)-\left(\frac{n^2+2 n+1^2}{4}\right) \\& =\frac{4 n^2+6 n+2-\left(3 n^2+6 n+3\right)}{12} \quad\left(\text { Since } \mathrm{a}^2-b^2=(a-b)(a+b)\right) \\& =\frac{n^2-1}{12} \\& =\frac{(n-1)(n+1)}{12}\end{aligned}

Therefore, the variance of uniform random variable X is,

V(X)=\frac{(n-1)(n+1)}{12}$$

Learn more about similar questions visit:

https://brainly.in/question/55165096?answering=false&answeringSource=feedQP

https://brainly.in/question/25639044?referrer=searchResults

#SPJ1

Answered by saikrupesh556
0

Answer:

Step-by-step explanation:

Similar questions