Question

For the following probability density function (p.d.f.) of X, find:
(i)$P(X < 1)$
(ii) $P(\left| X \right| < 1)$
If $\left\{ \begin{gathered}
  f(x) = \frac{{{x^2}}}{{18}},{\text{ }} - 3 < x < 3 \\
  {\text{ = }}0,{\text{ }}otherwise \\
\end{gathered} \right.$

Answer 1

Hint: We are given a probability density function and the probability of such functions is given by $\int\limits_a^b {f(x)dx} $ and in part one we are asked $P(X < 1)$and our function is defined at $ - 3 < x < 3$ so our limits will be -3 to 1 and it part two we asked to find $P(\left| X \right| < 1)$ , so our limits here will be -1 to 1.

Complete step-by-step answer:
We have
 .$
  f(x) = \frac{{{x^2}}}{{18}},{\text{ }} - 3 < x < 3 \\
  {\text{ = }}0,{\text{ }}otherwise \\
$.
(i)$P(X < 1)$
The value of .$P(X < 1)$.of a probability density function is given by the formula $\int\limits_a^b {f(x)dx} $
Here our function $f(x) = \frac{{{x^2}}}{{18}}$ is defined in the interval $ - 3 < x < 3$
Here we are asked for the probability of x less than 1
So ,
$
   \Rightarrow P(X < 1) = \int\limits_{ - 3}^1 {f(x)dx} \\
   \Rightarrow P(X < 1) = \int\limits_{ - 3}^1 {\frac{{{x^2}}}{{18}}dx} \\
$
Integrating using the formula $\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}}$
.$
   \Rightarrow P(X < 1) = \frac{1}{{18}}\left[ {\frac{{{x^3}}}{3}} \right]_{ - 3}^1 \\
   \Rightarrow P(X < 1) = \frac{1}{{18}}\left[ {\frac{{{{\left( 1 \right)}^3}}}{3} - \frac{{{{\left( { - 3} \right)}^3}}}{3}} \right] \\
   \Rightarrow P(X < 1) = \frac{1}{{18}}\left[ {\frac{1}{3} - \frac{{ - 27}}{3}} \right] \\
   \Rightarrow P(X < 1) = \frac{1}{{18}}\left[ {\frac{1}{3} + \frac{{27}}{3}} \right] = \frac{1}{{18}}\left[ {\frac{{28}}{3}} \right] = \frac{{28}}{{54}} = \frac{{14}}{{27}} \\
$.
(ii) .$P(\left| X \right| < 1)$.
Here we are given $\left| X \right| < 1$, hence the limit here varies between -1 and 1
So the value of $P(\left| X \right| < 1)$can be given by
$
   \Rightarrow P(\left| X \right| < 1) = \int\limits_{ - 1}^1 {f(x)dx} \\
   \Rightarrow P(\left| X \right| < 1) = \int\limits_{ - 1}^1 {\frac{{{x^2}}}{{18}}dx} \\
$
Integrating using the formula .$\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}}$.
$
   \Rightarrow P(\left| X \right| < 1) = \frac{1}{{18}}\left[ {\frac{{{x^3}}}{3}} \right]_{ - 1}^1 \\
   \Rightarrow P(\left| X \right| < 1) = \frac{1}{{18}}\left[ {\frac{{{{\left( 1 \right)}^3}}}{3} - \frac{{{{\left( { - 1} \right)}^3}}}{3}} \right] \\
   \Rightarrow P(\left| X \right| < 1) = \frac{1}{{18}}\left[ {\frac{1}{3} - \frac{{ - 1}}{3}} \right] \\
   \Rightarrow P(\left| X \right| < 1) = \frac{1}{{18}}\left[ {\frac{1}{3} + \frac{1}{3}} \right] = \frac{1}{{18}}\left[ {\frac{2}{3}} \right] = \frac{2}{{54}} = \frac{1}{{27}} \\
$
And hence we get our solution.

Note: Probability density function (PDF), in statistics, a function whose integral is calculated to find probabilities associated with a continuous random variable . Its graph is a curve above the horizontal axis that defines a total area, between itself and the axis, of 1. The percentage of this area included between any two values coincides with the probability that the outcome of an observation described by the probability density function falls between those values. Every random variable is associated with a probability density function