Continuous Random Variable Examples
In probability studies we come across many situations and apply various concepts to find the probability of an event. The numerical value of possible outcomes in a random experiment varies from trial to trial. Suppose you roll a fair die, the number of possible outcomes is $6$. That is, the die can roll on $1, 2, 3, 4, 5, 6$. If we define a variable $X, X$ can randomly vary from $1$ to $6$ in this experiment. But the outcomes are countable and hence the variable is called as discreet random variable. On the other hand if the continuous data can take any value within a range, then the variable is defined as continuous random variable. For example, if you study the heights of group of grown up persons, the variable can take any value including decimals within an appropriate range, says from $5$ ft. to $7$ ft.

Analysis of Continuous Random variables 

Let us now take a detailed study of continuous random variable. In a given range $[a, b]$ if the probability is equal for all the values of the random variable within the given interval, then the probability is uniformly distributed. It is also called as the rectangular distribution, the length of the rectangle is the range and the height is the probability of any value of the variable within the given range, say$p$. Clearly the area of such rectangle is $1$ as it represents the infinite sum of the probabilities of all possible outcomes.  Generalizing we can say it is a function represented as  $f(X) \{ X$ = $x \}$.

The following diagram helps us to understand better.

Analysis of Continuous Random variables

Now we get an interesting result! The length of the rectangle in the assumed range is $(b - a)$ and if the probability for any value of the variable is $p$,  then $p(b - a)$ = $1$. Therefore, $p$ = $\frac{1}{(b \times a)}$

We will now adopt a different approach. In case of uniformly distributed function $f(X)$ = $p$, where $p$ is a constant. The actual probability is $0$ at $x$ = $a$ and is $1$ at $x$ = $b$. Hence the probability cumulates from $0$ to $1$, in a linear manner in this case. If the interval width is the rate of variation or the slope is $\frac{1}{w}$. We can define a  cumulative density function shortly called as $CDF$) such that $C(X) \{X$ = $x \}$ = $kx$, where $k$ = $(\frac{1}{w})$.

In general, the probability distribution will not be uniform always and hence we cannot generalize $f(X)$ as a constant and may be in a general form as $f_X$(x),  The domain of  could be all real numbers, that is, $(-\infty$, $\infty)$, but in our probability study we can limit the interval as $[a, b]$. Since the probability is the area under the curve which represents the cumulative density function, we can say,                                                     

$P \{a < X \leq b\}$ = $F_x\ (b)\ -\ F_x$ (a) where $F_x$ is the integral of $f_X(x)$.

The most commonly known and important continuous distribution of a random variable is standard normal distribution, because many statistical data are normally distributed. Here the random variable is denoted as $z$ which is defined by the relation $z$ = $\frac{(\text{Random variable of the data\ -\ Mean of the data})}{\text{Standard deviation of the data}}$. The cumulative density function of a standard normal distribution is a bit complicated to integrate and evaluate. Instead the probability of occurrence represented by the random variable of the data can be directly found from a table called as z-score table.

Word Problems

Example 1: Govind is visiting his home village for a week. The parliamentary member of that constituency regularly spends a day in that village once in 60 days. What is the probability that Govind will meet the minister in his home village?

In a span of 60 days, the minister can visit on any of the days. In other words, each day the probability of ministers visit to the village is $\frac{1}{60}$. Hence, the cumulative density function of probability can be narrated as $C(X) \{X$ = $x\}$ = $(\frac{1}{60})$ $(x)$ 

Since Govind will be staying at the village for $7$ days the value of the random variable $x$ is $7$. Therefore, the probability of Govind meeting the minister is $C(X) \{X$ = $7\}$ = $(\frac{1}{60})$ $(7)$ = $\frac{7}{60}$.
Example 2:  A continuous random variable $X$ follows a normal distribution. What is $P(5 < X  < 12)$ ?

The width of the interval is $12 - 5$ = $7$ and hence $k$ = $\frac{1}{7}$

Therefore, the probability of the random variable for any value in the given range is $\frac{1}{7}$.

In other words, $P(5 < X  < 12)$ = $\frac{1}{7}$.
Example 3: The graph of probability density function of a random variable is shown below. What is the value of $k$?

Continuous Random Variable Examples

The graph describes the shape of a triangle and its area is given by $(\frac{1}{2})$ $\times\ k \times\18$ = $9k$

But this area represents the total probability of all possible values of the random variable which is $1$.

Therefore, $9k$ = $18$ and hence $k$ = $\frac{1}{9}$.
Example 4: A probability distribution function is defined as follows.

            $f_X(x)$ = $ke{-cx}$ for the interval [0, $\infty$)

a) Find the relation between$c$ and $k$

b) Find the cumulative density function assuming $c$ = $2$

c) Determine the probability for $3 < X < 5$.

a) The total probability for all positive values of the random variable is the area under the curve that represents $fX(x)$. In other words it is the definite integral of $fX(x)$ between the limits $0$ and infinity. But for the continuous random variable, the area under the graph of the function is $1$.

Therefore,  $\int_{0}^{\infty} ke^{-cx) }dx$ = $1$
 
 -$(\frac{k}{c})$ $[e^{-cx} ]_0^{\infty}$  = $1$

 -$(\frac{k}{c})$ $(0 - (1)$ = $1$
 
  $\frac{k}{c}$ = $1$ or $k$ = $c$.

b) If $c$ = $2$, then $k$ also = $2$ and hence $f_X (x)$ = $2e{-2x}$.

The cumulative density function is given by

$F_x(x)$ = $\int_{0}^{x}f_x(u)du$ where $0 \leq u < x$

$\int_{0}^{x} 2e^{-2u} du$ = $-[e^{-2u} ]_0^x $

Or, $F_X$(x)$ =  $1$  - $e^{-2x}$ 

c)  $P (3 < X < 5)$ = $F_X(5) - F_X (3)$ = $(1 - e^{-10})\ -\ (1 - e^{-6})$ = $e^{-6}\ -\ e^{-10}\ \approx\ 0.0024$
Example 5: The length of rod manufactured follows is a random variable follows a normal distribution with a mean length of $20$ cm and a standard division of $2$ cm. You randomly pick up a rod. What is the probability that  

a) the length of that rod is less than $19$ cm?

b) the length of the rod is between $21$ and $22$ cm?

a) The z-score for this case is $\frac{(19 \times 20)}{2}$ = $-0.5$

Therefore, the probability of the length of a rod less than $19$ cm, $P (X  < 19)$ is same as $P (Z < -0.5)$

Referring to the z-score table, the value corresponding to $-0.5$ is $0.3085$.

Hence the probability of the length of a randomly picked up rod less than $19$ cm is $0.3085$.

b) The z-score for a length of $21$ cm is $\frac{(21 \times 20)}{2}$ = $0.5$

    The z-score for a length of $22$ cm is $\frac{(22 \times 20)}{2}$ = 1

As per the z-score table, $P (Z$ = $1)$ = $0.8413$ and $P (Z$ = $0.5)$ = $0.6915$

So, $P (21 < X < 22)$ = $P (0.5 < Z < 1)$ = $P (Z$ = $1)\ -\ P (Z$ = $0.5)$ = $0.8413 - 0.6915$ = $0.1498$

Hence, the probability of the length of a randomly picked up rod between $21$ and $22$ cm is $0. 0.1498$