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# Continuous Random Variable

In probability theory, we study about continuous and discrete random variables. Continuous random variables differ from discrete random variables in that the data can take uncountable set of values and it is impossible to list all positive values in the sample space of a continuous variable.  For example, the age, weight or height of students in a class are all continuous random variables.

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## Definition

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If a variable can take on any value between its minimum value and its maximum value, it is called a continuous variable. In case of continuous random variable, we usually refer to the values in a particular interval and not at a point. Generally, continuous random variables represent the measured data.

## Mean of Continuous Random Variable

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A random variable may assume an infinite number of values within a given range. In the continuous random variable, the product of the value of the random variable and its probability density function is summed over all values of the random variable. The mean of a continuous random variable X with probability density function f(x) is given below.
The formula used for the mean value is   $E(X)$ = $\int_{-\infty}^{\infty}$$x f(x) dx ### Solved Example Question: Find the mean value for the continuous random variable f(x) = x, 0 \leq x \leq 2 . Solution: The given continuous function is f(x) = x, 0 \leq x \leq 2. The formula used for the mean value is E(X) = \int_ {- \infty}^{\infty}$$x f(x) dx$

$E(X)$ = $\int_{0}^{2}$$x \times x dx E(X) = \int_ {0}^{2}$$x^2 dx$

$E(X)$ = $[($$\frac{x^3}{3}$$)] _{0}^{2}$

$E(X)$ = $($$\frac{8}{3}$$) - (0)$

$E(X)$ = $\frac{8}{3}$

The mean value for continuous random variable f(x) = x, 0 $\leq$ x $\leq$ 2 is $\frac{8}{3}$.

## Examples

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Given below are some of the examples on continuous random variable.

### Solved Examples

Question 1: A continuous random variable X has the following probability function

 Value of X (x) 0 1 2 3 4 5 6 7 p(x) 0 k 2k 2k 3k k$^2$ 2k$^2$ 7k$^2$+k

(i)     Find k
(ii)    Evaluate P(x < 6), P(x $\geq$ 6) and P(0 < x < 5)
(iii)    If P(X $\leq$ x) > $\frac{1}{2}$, find the minimum value of x.

Solution:

Step 1:
$\sum _{x\rightarrow 0}^{7}$$p(x) = 1, we have k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1 \Rightarrow 10k^2 + 9k = 1 \Rightarrow 10k^2 + 9k - 1 = 0 \Rightarrow (10k - 1)(k + 1) = 0 \Rightarrow 10k - 1 = 0 and k + 1 = 0 \Rightarrow k = \frac{1}{10} and k = -1. Step 2: Here k = - 1 is discarded as p(x) cannot be negative. P(x < 6) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) = 0 + k + 2k + 2k + 3k + k^2 = k^2 + 8k = (\frac{1}{10})^2 + 8(\frac{1}{10}) = \frac{81}{100} Step 3: P(x \geq 6) = 1 - P(x < 6) = 1 - \frac{81}{100} = \frac{19}{100} Step 4: P(0 < x < 5) = P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) = k + 2k + 2k + 3k = 8k = \frac{8}{10} By trial, P(X \leq x) > \frac{1}{2}, We have x = 4 which is the minimum value of x. Question 2: Two coins are tossed simultaneously. Getting a head is termed as success. Find the probability distribution of the number of successes (heads). Solution: In this case the continuous random variable X denotes the number of heads. Step 1: The sample space, S= {HH, HT, TH,TT} Range of X= {0,1, 2} Step 2: Now, P(X = 0) = Probability of no heads, that is P(TT) = \frac{1}{2}$$\times$$\frac{1}{2} = \frac{1}{4} P(X = 1) = Probability of one head, that is P (HT, TH) = \frac{1}{4}+\frac{1}{4} = \frac{1}{2} Step 3: P(X = 2) = Probability of two heads, that is P (HH) = \frac{1}{2}$$\times$$\frac{1}{2}$

= $\frac{1}{4}$

So the probability distribution of the continuous random variable is

 X = xi 0 1 2 P(X = xi) $\frac{1}{4}$ $\frac{1}{2}$ $\frac{1}{4}$

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