A random variable is continuous if the values can be expressed as an interval. Calculations are for continuous variables, for example heights or times, that is, something that can be measured rather than counted. In order to calculate probabilities, we need to find the area under graphs, so it is important that you are confident with integration techniques before working through this page. In the exam, questions are usually the long part B questions. Typical questions are given in the exam-style questions below.
On this page, you should learn about
continuous random variables and their probability density functions mode of continuous random variables median of continuous random variables mean of continuous random variables variance of continuous random variables Here is a quiz that gets you to check the initial conditions for continuous random variables
START QUIZ! The graph below shows a continuous random variable, X
Find the value of a
The area under the graph equals 1
5a = 1
a = \(\frac{1}{5}\)
Check
The graph below shows a continuous random variable, X
Find the value of a
The area under the graph equals 1
\(\frac{4a}{2}=1\)
a = \(\frac{1}{2}\)
Check
The graph below shows a continuous random variable, X, where the probability density function is given by
\(f\left(x\right)=\left\{\begin{matrix}\frac{x^2}{9}\\0\\\end{matrix}\ \ \ \ \ \ \right.\begin{matrix}0\le x<a\\otherwise\\\end{matrix}\)
Find the value of a
The area under the graph equals 1
\(\int_{0}^{a}{\frac{x^2}{9}dx=1}\)
\(\left[\frac{x^3}{27}\right]_0^a=1\)
\(\frac{a^3}{27}=1\)
\(a^3=27\)
a = 3
Check
The graph below shows a continuous random variable, X
Find the value of a
The area under the graph equals 1
\(a^2+\frac{a^2}{2}=1\)
\(\frac{3a^2}{2}=1\)
\(a^2=\frac{2}{3}\)
\(a=\frac{\sqrt{2}}{\sqrt{3}}\)
Rationalise the denominator
\(a=\frac{\sqrt{2}\sqrt{3}}{3}\)
\(a=\frac{\sqrt{6}}{3}\)
Check
The probability density function of X is given by
\(f\left(x\right)=\left\{\begin{matrix}0.5\\0\\\end{matrix}\ \ \ \ \ \ \right.\begin{matrix}0\le x<a\\otherwise\\\end{matrix}\)
Find the value of a that makes this a continuous random variable
The area under the graph of the function equals 1
0.5 x 2 = 1
Check
The probability density function of X is given by
\(f\left(x\right)=\left\{\begin{matrix}\frac{x}{8}\\0\\\end{matrix}\ \ \ \ \ \ \right.\begin{matrix}0\le x<a\\otherwise\\\end{matrix}\)
Find the value of a that makes this a continuous random variable
The area under the graph of the function equals 1
There is no need to use integration. We can find the area of the triangle
\(\frac{a^2}{16}=1\)
a² = 16
a = 4
Check
The probability density function of X is given by
\(f\left(x\right)=\left\{\begin{matrix}e^x\\0\\\end{matrix}\ \ \ \ \ \ \right.\begin{matrix}0\le x<lna\\otherwise\\\end{matrix}\)
Find the value of a that makes this a continuous random variable
The area under the graph of the function equals 1
\(\int_{0}^{lna}{e^xdx=1}\)
\(\left[e^x\right]_0^{lna}=1\)
\(e^{lna}-e^0=1\)
a - 1 = 1
a = 2
Check
The probability density function of X is given by
\(f\left(x\right)=\left\{\begin{matrix}\frac{1}{6}x\\0\\\end{matrix}\ \ \ \ \ \ \right.\begin{matrix}a\le x<4\\otherwise\\\end{matrix}\)
Find the value of a that makes this a continuous random variable
The area under the graph of the function equals 1
We do not need to use integration to solve this
\(\frac{4\times\frac{4}{6}}{2}-\frac{a\times\frac{a}{6}}{2}=1\)
\(\frac{16-a^2}{12}=1\)
16 - a² = 12
a² = 4
a = 2
Check
The graph below shows a continuous random variable, X
Find \(P(1\le X<1.5)\)
We find the area under the graph between 1 and 1.5
Check
The graph below shows a continuous random variable, X
Find \(P(0.5\le X<1.5)\)
We find the area under the graph between 0.5 and 1.5
Area \(\frac{1.5\times 0.75}{2}-\frac{0.5\times 0.25}{2}=0.5\)
Check
Here is a quiz about finding the mean, mode and median from a continuous random variable
START QUIZ! The graph below shows the continuous random variable X.
Find E(X)
E(X) is the mean.
Since this is a symmetrical distribution, there is no need to use any integration. It is the mid-value.
Check
The graph below shows the continuous random variable X.
Find E(X)
E(X) is the mean.
Since this is a symmetrical distribution, there is no need to use any integration. It is the mid-value.
Check
The graph below shows the continuous random variable X.
Find the median.
Check
The graph below shows the continuous random variable X.
Find the mode
Mode is the value at which the probability density function reaches a maximum
Check
The graph below shows the continuous random variable X.
Find the mode
Mode is the value at which the probability density function reaches a maximum
Check
The graph below shows the continuous random variable X.
Find the median
Check
The graph below shows the continuous random variable X.
Find the mode
Mode is the value at which the probability density function reaches a maximum
Check
The graph below shows the continuous random variable X.
Find the median
Let m = median
\(\frac{m\times \frac{m}{2}}{2}=\frac{1}{2}\)
m² = 2
\(m=\sqrt{2}\)
Check
The graph below shows the continuous random variable X where
\(f\left(x\right)=\left\{\begin{matrix}\frac{x}{2}\\0\\\end{matrix}\ \ \ \ \ \ \ \right.\begin{matrix}0\le x<2\\otherwise\\\end{matrix}\)
E(X) = \(\frac{a}{b}\) where \(\frac{a}{b}\) is a fraction in its simplest terms.
Find a and b
\(E\left(X\right)=\int_{0}^{2}{\frac{x}{2}\bullet x d x}\)
\(E\left(X\right)=\int_{0}^{2}{\frac{x^2}{2} d x}\)
\(E\left(X\right)=\left[\frac{x^3}{6}\right]_0^2\)
\(E(X)=\frac{8}{6}=\frac{4}{3}\)
Check
The continuous random variable X is given by the following probability density function
\(f\left(x\right)=\left\{\begin{matrix}1-x\\0\\\end{matrix}\ \ \ \ \ \ \ \right.\begin{matrix}0\le x<1\\otherwise\\\end{matrix}\)
E(X) = \(\frac{a}{b}\) where \(\frac{a}{b}\) is a fraction in its simplest terms.
Find a and b
\(E\left(X\right)=\int_{0}^{1}{(1-x) x d x}\)
\(E\left(X\right)=\int_{0}^{1}{(x-x^2) d x}\)
\(E\left(X\right)=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1\)
\(E(X)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
Check
The probability density function of X is given by
\(f\left(x\right)=\left\{\begin{matrix}ax^n\\0\\\end{matrix}\ \ \ \ \ \ \ \right.\begin{matrix}0\le x<1\\otherwise\\\end{matrix}\)
a) Show that a = n + 1
b) Given that E(X) = 0.75, find a and n
Hint For continuous random variables, the total area under the function = 1
\(\int_{-\infty}^{\infty}{f\left(x\right)dx=1}\)
Full Solution
The continuous random variable X has a probability density function given by
\(f\left(x\right)=\left\{\begin{matrix}a\\asin\left(\frac{\pi x}{4}\right)\\0\\\end{matrix}\ \ \ \ \ \ \ \right.\begin{matrix}0\le x<2\\2\le x<4\\otherwise\\\end{matrix}\)
a) Draw a sketch of f
b) Show that the value of \(a=\frac{\pi}{2\pi+4}\)
c) Find E(X)
d) Find the exact value of the median of X
e) Find \(P(X\le2|X\le3)\)
Hint b) The area under the graph of f equals 1 (sum of probabilities=1). Use this fact to find a
c) \(E\left(X\right)=\int_{-\infty}^{\infty}xf\left(x\right)dx\)
d) \(\int_{-\infty}^{m}{f\left(x\right)dx=0.5}\)
From the graph it is obvious that the median occurs between \(0 \le x<2\) , so there is no need to use Calculus
e) \(P(X\le2|X\le3)=\frac{P(X\le2)}{P(X\le3)}\)
Full Solution
The continuous random variable X has a probability density function given by
\(f\left(x\right)=\left\{\begin{matrix}k\bullet a r c o s\left(x\right)\\0\\\end{matrix}\ \ \ \ \ \ \ \right.\begin{matrix}-1\le x<1\\otherwise\\\end{matrix}\)
a) Draw a sketch of f
b) State the mode of X
c) Find \(\int a r c o s\left(x\right)dx\)
d) Find E(X)
e) Find Var(X)
Hint c) You will need to use Integration by parts and make the 'other part' 1
\(\int{1\bullet a r c o s\left(x\right)dx}\)
d) and e) You can use your graphical calculator to work these out.
Full Solution MY PROGRESS
Self-assessment How much of Continuous Random Variables have you understood?
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