$ X $ is the random variable that can take any one of the values, $0, 1, 7, 11$ and $12$. The probability mass function for $ X $ is $ P(X = 0) = 0.4; P(X = 1) = 0.3; P(X = 7) = 0.1; $ $ P(X = 11) = 0.1; P(X = 12) = 0.1 $ Then, the variance of $ X $ is
Calculating Random Variable Variance
The question requires calculating the variance of a discrete random variable $X$. The variance $Var(X)$ is defined as $Var(X) = E[X^2] - (E[X])^2$. We first need to determine the expected value $E[X]$ and the expected value of $X^2$, denoted $E[X^2]$.
Variance Calculation Steps
Step 1: Calculate Expected Value, $E[X]$
The expected value $E[X]$ is found by summing the product of each possible value of $X$ and its corresponding probability $P(X=x)$.
The formula is: $ E[X] = \sum x \cdot P(X=x) $
Using the provided probability mass function:
| Value ($x$) | Probability ($P(X=x)$) | $x \cdot P(X=x)$ |
| 0 | 0.4 | $0 \times 0.4 = 0$ |
| 1 | 0.3 | $1 \times 0.3 = 0.3$ |
| 7 | 0.1 | $7 \times 0.1 = 0.7$ |
| 11 | 0.1 | $11 \times 0.1 = 1.1$ |
| 12 | 0.1 | $12 \times 0.1 = 1.2$ |
| Sum | $3.3$ |
Thus, the expected value is $ E[X] = 3.3 $.
Step 2: Calculate Expected Value of $X^2$, $E[X^2]$
The expected value of $X^2$ is calculated similarly, by summing the product of the square of each value of $X$ and its corresponding probability.
The formula is: $ E[X^2] = \sum x^2 \cdot P(X=x) $
Calculating the squared values and their products:
| Value ($x$) | $x^2$ | Probability ($P(X=x)$) | $x^2 \cdot P(X=x)$ |
| 0 | $0^2 = 0$ | 0.4 | $0 \times 0.4 = 0$ |
| 1 | $1^2 = 1$ | 0.3 | $1 \times 0.3 = 0.3$ |
| 7 | $7^2 = 49$ | 0.1 | $49 \times 0.1 = 4.9$ |
| 11 | $11^2 = 121$ | 0.1 | $121 \times 0.1 = 12.1$ |
| 12 | $12^2 = 144$ | 0.1 | $144 \times 0.1 = 14.4$ |
| Sum | $31.7$ |
So, the expected value of $X^2$ is $ E[X^2] = 31.7 $.
Step 3: Calculate Variance, $Var(X)$
Finally, apply the variance formula using the computed values of $E[X]$ and $E[X^2]$.
$ Var(X) = E[X^2] - (E[X])^2 $
$ Var(X) = 31.7 - (3.3)^2 $
$ Var(X) = 31.7 - 10.89 $
$ Var(X) = 20.81 $
Final Answer
The variance of the random variable $X$ is $20.81$.
Important Questions from Random Variables Basics
The length of time X, needed by an examinee of competition to complete a 1-hour exam, is a random variable with
PDF \(f(x)=\dfrac{6}{5}(x^2+x);0 \le x \le 1.\) , The value of F(0.5) is:If X follows a binomial distribution with n = 6 and \(p=\dfrac{1}{4}\) then the skewness of X is:
If the customers arrive in a shop in Poisson fashion with parameter λ, the fourth raw moment \(\mu_4^{'}\) for the inter-arrival time is:
A discrete random variable X has the probability functions as:
X
0
1
2
3
4
5
6
7
8
f(x)
K
2k
3k
5k
5k
4k
3k
2k
k
The value of E(X) is:What percentage of scores falls within three standard deviations from the mean for the normal variate?
