A continuous random variable x has a probability density function \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {6x\left( {1 - x} \right),}&{0 < x \le 1}\\ {0,}&{otherwise} \end{array}} \right.\) Then the variance of x is:
Finding Variance of a Continuous Random Variable
To find the variance of a continuous random variable \(x\) with a given probability density function (PDF), \(f(x)\), we use the formula:
\(Var(x) = E[x^2] - (E[x])^2\)
where \(E[x]\) is the expected value (mean) of \(x\) and \(E[x^2]\) is the expected value of \(x^2\).
The given probability density function is:
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {6x\left( {1 - x} \right),}&{0 < x \le 1}\\ {0,}&{otherwise} \end{array}} \right.\)
Step-by-Step Calculation of Expected Value E[x]
The expected value \(E[x]\) for a continuous random variable is calculated using the integral:
\(E[x] = \int_{-\infty}^{\infty} x f(x) dx\)
Given the PDF is non-zero only for \(0 < x \le 1\), the integral becomes:
\(E[x] = \int_{0}^{1} x \cdot 6x(1 - x) dx\)
First, expand the expression inside the integral:
\(x \cdot 6x(1 - x) = 6x^2(1 - x) = 6x^2 - 6x^3\)
Now, integrate this expression from 0 to 1:
\(E[x] = \int_{0}^{1} (6x^2 - 6x^3) dx\)
Integrate term by term:
\(\int (6x^2 - 6x^3) dx = 6 \int x^2 dx - 6 \int x^3 dx = 6 \frac{x^3}{3} - 6 \frac{x^4}{4} = 2x^3 - \frac{3}{2}x^4\)
Evaluate the definite integral from 0 to 1:
\(E[x] = \left[ 2x^3 - \frac{3}{2}x^4 \right]_{0}^{1}\)
\(E[x] = \left( 2(1)^3 - \frac{3}{2}(1)^4 \right) - \left( 2(0)^3 - \frac{3}{2}(0)^4 \right)\)
\(E[x] = \left( 2 - \frac{3}{2} \right) - (0 - 0)\)
\(E[x] = \frac{4}{2} - \frac{3}{2} = \frac{1}{2}\)
So, the expected value \(E[x]\) is \(\frac{1}{2}\).
Step-by-Step Calculation of Expected Value E[x<sup>2</sup>]
The expected value \(E[x^2]\) for a continuous random variable is calculated using the integral:
\(E[x^2] = \int_{-\infty}^{\infty} x^2 f(x) dx\)
Given the PDF is non-zero only for \(0 < x \le 1\), the integral becomes:
\(E[x^2] = \int_{0}^{1} x^2 \cdot 6x(1 - x) dx\)
First, expand the expression inside the integral:
\(x^2 \cdot 6x(1 - x) = 6x^3(1 - x) = 6x^3 - 6x^4\)
Now, integrate this expression from 0 to 1:
\(E[x^2] = \int_{0}^{1} (6x^3 - 6x^4) dx\)
Integrate term by term:
\(\int (6x^3 - 6x^4) dx = 6 \int x^3 dx - 6 \int x^4 dx = 6 \frac{x^4}{4} - 6 \frac{x^5}{5} = \frac{3}{2}x^4 - \frac{6}{5}x^5\)
Evaluate the definite integral from 0 to 1:
\(E[x^2] = \left[ \frac{3}{2}x^4 - \frac{6}{5}x^5 \right]_{0}^{1}\)
\(E[x^2] = \left( \frac{3}{2}(1)^4 - \frac{6}{5}(1)^5 \right) - \left( \frac{3}{2}(0)^4 - \frac{6}{5}(0)^5 \right)\)
\(E[x^2] = \left( \frac{3}{2} - \frac{6}{5} \right) - (0 - 0)\)
To subtract the fractions, find a common denominator, which is 10:
\(E[x^2] = \frac{3 \cdot 5}{2 \cdot 5} - \frac{6 \cdot 2}{5 \cdot 2} = \frac{15}{10} - \frac{12}{10} = \frac{15 - 12}{10} = \frac{3}{10}\)
So, the expected value \(E[x^2]\) is \(\frac{3}{10}\).
Calculating the Variance Var(x)
Now we use the formula for variance:
\(Var(x) = E[x^2] - (E[x])^2\)
Substitute the calculated values for \(E[x]\) and \(E[x^2]\):
\(Var(x) = \frac{3}{10} - \left(\frac{1}{2}\right)^2\)
Calculate \(\left(\frac{1}{2}\right)^2\):
\(\left(\frac{1}{2}\right)^2 = \frac{1^2}{2^2} = \frac{1}{4}\)
Substitute this back into the variance formula:
\(Var(x) = \frac{3}{10} - \frac{1}{4}\)
To subtract these fractions, find a common denominator, which is 20:
\(Var(x) = \frac{3 \cdot 2}{10 \cdot 2} - \frac{1 \cdot 5}{4 \cdot 5}\)
\(Var(x) = \frac{6}{20} - \frac{5}{20}\)
\(Var(x) = \frac{6 - 5}{20} = \frac{1}{20}\)
The variance of the continuous random variable \(x\) is \(\frac{1}{20}\).
Let's compare this result with the given options:
- Option 1: \(\dfrac{1}{20}\)
- Option 2: \(\dfrac{1}{40}\)
- Option 3: \(\dfrac{1}{5}\)
- Option 4: \(\dfrac{1}{2}\)
Our calculated variance matches Option 1.
Summary of Variance Calculation Steps
| Step | Description | Formula / Calculation |
|---|---|---|
| 1 | Identify the PDF, \(f(x)\). | \(f(x) = 6x(1-x)\) for \(0 < x \le 1\) |
| 2 | Calculate the expected value \(E[x]\). | \(E[x] = \int_{0}^{1} x f(x) dx = \int_{0}^{1} (6x^2 - 6x^3) dx = \frac{1}{2}\) |
| 3 | Calculate the expected value \(E[x^2]\). | \(E[x^2] = \int_{0}^{1} x^2 f(x) dx = \int_{0}^{1} (6x^3 - 6x^4) dx = \frac{3}{10}\) |
| 4 | Calculate the variance \(Var(x)\). | \(Var(x) = E[x^2] - (E[x])^2 = \frac{3}{10} - \left(\frac{1}{2}\right)^2 = \frac{1}{20}\) |
Revision Table: Key Concepts in Variance and Expected Value
| Concept | Definition | Formula (Continuous RV) |
|---|---|---|
| Expected Value (Mean) | The average value of a random variable over many trials. | \(E[x] = \int_{-\infty}^{\infty} x f(x) dx\) |
| Expected Value of g(x) | The average value of a function of a random variable. | \(E[g(x)] = \int_{-\infty}^{\infty} g(x) f(x) dx\) |
| Variance | A measure of how spread out the values of a random variable are from its mean. | \(Var(x) = E[(x - E[x])^2] = E[x^2] - (E[x])^2\) |
| Probability Density Function (PDF) | A function describing the relative likelihood for a continuous random variable to take on a given value. The integral of the PDF over its support must equal 1. | \(\int_{-\infty}^{\infty} f(x) dx = 1\) |
Additional Information: Properties of Variance
- The variance of a constant \(c\) is zero: \(Var(c) = 0\).
- The variance of a linear transformation \(ax + b\) is: \(Var(ax + b) = a^2 Var(x)\). Note that adding a constant \(b\) does not change the variance.
- Variance is always non-negative: \(Var(x) \ge 0\).
- The square root of the variance is the standard deviation: \(SD(x) = \sqrt{Var(x)}\).
Important Questions from Mean and Variance of Random variables
X is a non-negative integer valued random variable with
\( P(X = x) =\left\{ \begin{matrix} \dfrac {x+1}{2^{(x+2)}} & x = 0, 1, 2... \\\ 0 & \rm{otherwise} \end{matrix} \right.\)
Then, mean and variance of X are respectively
In eight throws of a die, 5 or 6 is considered a success. The mean and standard deviation of total number of successes is respectively given by
Out of 100 numbers, 20 are 4's, 40 are 5's, 30 are 6's, and the remaining are 7's. The arithmetic mean of the numbers is:
The median of the normal distribution with mean and variance \(\mu\) and \(\sigma^2\) is:
Let a continuous random variable \( X \) have probability density function (pdf):
\[f(x) = \begin{cases} -0.75 \, x^2 + 1.5x & \text{for } 0 < x < 2 \\ 0, & \text{otherwise} \end{cases}\]
Find the mode of \( X \).
