Let $Y = Z^2$, $Z = \frac{X - \mu}{\sigma}$, where $X$ is a normal random variable with mean $\mu$ and variance $\sigma^2$. The variance of $Y$ is
Variance Calculation for Y = Z^2
We are given a normal random variable $X$ with mean $\mu$ and variance $\sigma^2$, denoted as $X \sim N(\mu, \sigma^2)$.
A standardized variable $Z$ is defined as $Z = \frac{X - \mu}{\sigma}$. This transformation makes $Z$ a standard normal random variable, meaning $Z \sim N(0, 1)$.
We need to find the variance of $Y = Z^2$.
Properties of Standard Normal Variable Z
- For $Z \sim N(0, 1)$, the expected value is $E[Z] = 0$.
- The variance is $Var(Z) = E[Z^2] - (E[Z])^2 = 1$.
- Since $E[Z] = 0$, we have $E[Z^2] - 0^2 = 1$, which means $E[Z^2] = 1$.
Relationship between Y and Chi-Squared Distribution
The variable $Y$ is defined as $Y = Z^2$. Since $Z \sim N(0, 1)$, the square of $Z$, which is $Z^2$, follows a Chi-squared distribution with 1 degree of freedom. This is denoted as $Y \sim \chi^2(1)$.
Calculating Variance of Y
The variance of a Chi-squared distribution with $k$ degrees of freedom ($\chi^2(k)$) is given by the formula $Var = 2k$.
In our case, $Y \sim \chi^2(1)$, so $k=1$. Applying the formula:
$ Var(Y) = 2k = 2 \times 1 = 2 $
Therefore, the variance of $Y$ is 2.
Important Questions from Random Variables
Consider the following probability density function for a random variable x.
(a) \(f_{1}(x)=1\ ;\ -1\le x\le 1\)
(b) \(f_{2}(x)=x\ ;\ 0\le x\le 1\)
(c) \(f_{3}(x)=(1-x)\ ;\ -1\le x\le 1\)
Arrange the above functions in terms of the increasing value of mean of random variable x.
Considering all the symbols with their usual meanings, match the following :
List - I List - II (a) \(f_{X}(x)\) (i) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx\) (b) \(\displaystyle\int_{-\infty}^{\infty}f_{X}(x)\,dx\) (ii) \(\displaystyle\int_{-\infty}^{a}f_{X}(x)\,dx\) (c) \(F_{X}(a)\) (iii) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\) (d) \(f_{Y}(y)\) (iv) 1 Codes :
The density function of a random variable is given by \(p(x)=Ke^{-\frac{x^{2}}{2}}\) for \(-\infty \lt x \lt \infty\). The value of K should be :
If the odds in favour of any random event A are 5 ∶ 6, then the odds against the event are:
If random variable X follows binomial distribution with parameter n and p with mean 15 and variance 10, then the value of mode is
