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
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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
a =
The area under the graph equals 1
5a = 1
a = \(\frac{1}{5}\)
The graph below shows a continuous random variable, X
Find the value of a
a =
The area under the graph equals 1
\(\frac{4a}{2}=1\)
a = \(\frac{1}{2}\)
The graph below shows a continuous random variable, X, where the probability density function is given by
Find the value of a
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
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
\(a=\frac{\sqrt{2}\sqrt{3}}{3}\)
\(a=\frac{\sqrt{6}}{3}\)
The probability density function of X is given by
Find the value of a that makes this a continuous random variable
a =
The area under the graph of the function equals 1
0.5 x 2 = 1
The probability density function of X is given by
Find the value of a that makes this a continuous random variable
a =
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
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
a =
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
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
a =
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
The graph below shows a continuous random variable, X
Find \(P(1\le X<1.5)\)
probability =
We find the area under the graph between 1 and 1.5
The graph below shows a continuous random variable, X
Find \(P(0.5\le X<1.5)\)
probability =
We find the area under the graph between 0.5 and 1.5
Area \(\frac{1.5\times 0.75}{2}-\frac{0.5\times 0.25}{2}=0.5\)
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) =
E(X) is the mean.
Since this is a symmetrical distribution, there is no need to use any integration. It is the mid-value.
The graph below shows the continuous random variable X.
Find E(X)
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.
The graph below shows the continuous random variable X.
Find the median.
Median =
It is the middle value
The graph below shows the continuous random variable X.
Find the mode
Mode =
Mode is the value at which the probability density function reaches a maximum
The graph below shows the continuous random variable X.
Find the mode
Mode =
Mode is the value at which the probability density function reaches a maximum
The graph below shows the continuous random variable X.
Find the median
Median =
It is the middle value
The graph below shows the continuous random variable X.
Find the mode
Mode =
Mode is the value at which the probability density function reaches a maximum
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
\(m=\sqrt{2}\)
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
a =
b =
\(E\left(X\right)=\int_{0}^{2}{\frac{x}{2}\bullet x d x}\)
\(E\left(X\right)=\int_{0}^{2}{\frac{x^2}{2} d x}\)
\(E\left(X\right)=\left[\frac{x^3}{6}\right]_0^2\)
\(E(X)=\frac{8}{6}=\frac{4}{3}\)
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
a =
b =
\(E\left(X\right)=\int_{0}^{1}{(1-x) x d x}\)
\(E\left(X\right)=\int_{0}^{1}{(x-x^2) d x}\)
\(E\left(X\right)=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1\)
\(E(X)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
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
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
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
Full Solution
How much of Continuous Random Variables have you understood?
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