In another chapter, we introduced the idea of a probability density function belonging to a continuous random variable.
A quantity that is to be measured and that can have any value within a given range may give rise to a continuous random variable. However, if we ask, What is the probability that the random variable takes precisely some particular value $a$a? the answer has to be zero. This is because there are infinitely many values the random variable could take and if each of them had a non-zero probability, however small, the total probability of all the possible outcomes would be greater than $1$1, which is impossible.
Instead, we ask, What is the probability that the random variable takes a value within some specified interval? Then, however small the interval, as long as it is greater than zero it will be possible to assign a definite non-zero probability to the event corresponding to the random variable being within the interval.
A probability density function is called a PDF for short.
We also define a cumulative distribution function, CDF. The height of the CDF graph answers the question, What is the probability that the random variable $X$X takes a value less than $x$x? That is, at a point $x$x on the horizontal axis, the CDF gives the probability of the event $X
A CDF is always a non-decreasing function. It has the value $0$0 at or below the minimum value of $X$X and rises to $1$1 at the maximum value of $X$X.
In terms of an integral, the CDF corresponding to a density function $f$f is $F(x)=\int_{-\infty}^xf(t)\mathrm{d}t$F(x)=∫x−∞f(t)dt
Example 1
A cable of length $100$100 m is stretched to breaking point. It is thought that the point at which the cable breaks is equally likely to be anywhere along its length. What probability density function models the likely breaking point of the cable? What is the corresponding cumulative distribution function?
Given any interval in the length of the cable, we would say that the probability of the break occurring in that interval is proportional to the length of the interval. If we define a random variable $X$X to be the distance from the left-hand end of the cable to the position of the break, we can construct the PDF
$f(x)=\frac{1}{100}$f(x)=1100
This is called a uniform PDF. We would expect, for example, that it would be just as likely for the break to occur between $0$0 and $25$25 m as between $50$50 m and $75$75 m or in any other interval of length $25$25 m. So, for the break to occur between, say, $35$35 m and $60$60 m the probability given by this PDF is $\frac{1}{100}(60-35)=\frac{1}{4}$1100(60−35)=14.
The corresponding CDF is
$F(x)=0$F(x)=0 for $x<0$x<0
$F(x)=\int_0^x\frac{1}{100}\mathrm{d}t=\frac{x}{100}$F(x)=∫x01100dt=x100 for $0\le x\le100$0≤x≤100
$F(x)=1$F(x)=1 for $x>100$x>100.
Worked Examples
Question 1
Consider the probability density function $p\left(x\right)$p(x) drawn below for a random variable $X$X.
Calculate the area between $p(x)$p(x) and the $x$x axis.
Which feature(s) of $p\left(x\right)$p(x) is also a feature of all probability distribution functions? Select all options that apply.
$p\left(x\right)$p(x) is positive for all values of $x$x.
A$p\left(x\right)$p(x) is defined in the region $-\infty
B$p\left(x\right)$p(x) is only defined in the region $10\le x\le80$10≤x≤80.
CThe area under $p\left(x\right)$p(x) is equal to $1$1.
DCalculate $P$P$($($X$X$\le$≤$54$54$)$) using geometric reasoning.
Calculate $P$P$($($X$X$>$>$34$34$)$) using geometric reasoning.
Calculate $P$P$($($44$44$<$<$X$X$\le$≤$53$53$)$) using geometric reasoning.
Calculate $P$P$($($X$X$\le$≤$56$56$\mid$∣$X\ge44$X≥44$)$) using geometric reasoning.
Question 2
Consider the probability density function $p\left(x\right)$p(x) drawn below for a random variable $X$X.
Calculate the area between $p(x)$p(x) and the $x$x axis, showing working.
Which features of $p\left(x\right)$p(x) are also features of all probability distribution functions? Select every option that applies.
$p\left(x\right)$p(x) is defined in the region $-\infty
AThe total area under $p\left(x\right)$p(x) is equal to $1$1.
BThe total area under $p\left(x\right)$p(x) is less than $1$1.
C$p\left(x\right)$p(x) is $0$0 for $x$x values less than $0$0.
DState the probability density function for $X$X.
$\editable{}$; $0\le x\le3$0≤x≤3 $p\left(x\right)$p(x)$=$= $\editable{}$; $3 $\editable{}$; otherwise Calculate $P($P($X\ge4$X≥4$)$).
Using your answers from parts (c)-(d), calculate $P($P($X>3$X>3$|$|$X<4$X<4$)$). Leave your answer as a fully simplified fraction.
Question 3
The probability density function of a random variable $X$X is defined by $p\left(x\right)=kx^2$p(x)=kx2 for $0\le x\le\sqrt[3]{4}$0≤x≤3√4, and $p\left(x\right)=0$p(x)=0 elsewhere, and is drawn below. Note that $k$k is a constant.
Use calculus techniques to determine the value for $k$k.