A continuous random variable $X$ has probability density function given by
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{k}{x}\,\,\,\,\,\,1 \leqslant x \leqslant 9} \\
  {0\,\,\,\,\,\,x < 1\,or\,x > 9}
\end{array}} \right.\]
a. Find the value of $k.$
b. Find the mean and variance of $X$, giving your answer correct to three decimal places.

Hint: In this question, we are going to find the value of $k$ and then to find mean and variance of $X$.
First we are going to find the value of $k$ by using the probability density function.
Next we are going to find the mean and variance of $X$ for the probability density function.
Hence we can get the required solution.

Formula used: The probability density function is written in the form
$\int\limits_{ - \infty }^\infty {f\left( x \right)dx = 1} $
Mean of the probability density function is written as
$\mu = E\left( X \right) = \int\limits_{ - \infty }^\infty {xf\left( x \right)dx} $
Variance of the probability density function is written as
${\sigma ^2} = Var\left( X \right) = E\left( {{X^2}} \right) - {\mu ^2}$
${\sigma ^2} = \int\limits_{ - \infty }^\infty {{x^2}f\left( x \right)dx - {\mu ^2}} $

Complete step by step solution:
In this question, we are going to find the value of $k$ and then find mean and variance of $X$ .
a.)First we are going to find the value of $k$ by using the probability density function.
Since $f\left( x \right)$is a probability density function$\int\limits_{ - \infty }^\infty {f\left( x \right)dx = 1} $
That is $\int\limits_1^9 {f\left( x \right)dx = 1} $
Substitute the value of $f\left( x \right)$
$ \Rightarrow \int\limits_1^9 {\dfrac{k}{x}dx = 1} $
$k$is a constant and it can be taken outside
$ \Rightarrow k\int\limits_1^9 {\dfrac{1}{x}dx = 1} $
Integration of $\dfrac{1}{x}dx$ is $\ln \left| x \right| + c$
$ \Rightarrow k\left[ {\ln \left| x \right| + C} \right]_1^9 = 1$
Substitute the value of upper and lower limit in $x,$ that is upper limit minus lower limit, we get
$ \Rightarrow k\left[ {\left( {\ln \left| 9 \right| + C} \right) - \left( {\ln \left| 1 \right| + C} \right)} \right] = 1$
$ \Rightarrow k\left[ {2.19772} \right] = 1$
$ \Rightarrow k = \dfrac{1}{{2.19772}}$
$ \Rightarrow k = 0.45501$
Hence we get the value of $k$ as $0.45501$
b.) now we are going to find the mean and variance of the probability density function.
$\mu = \int\limits_{ - \infty }^\infty {xf\left( x \right)dx} $
Substitute the value of $f\left( x \right)$
$ \Rightarrow \mu = \int\limits_1^9 {x \times \dfrac{k}{x}dx} $
$ \Rightarrow \mu = \int\limits_1^9 {kdx} $
$k$is a constant and it can be taken outside
$ \Rightarrow \mu = k\int\limits_1^9 {dx} $
$ \Rightarrow \mu = k\left[ {x + C} \right]_1^9$
Substitute the value of upper and lower limit in $x,$ that is upper limit minus lower limit, we get
$ \Rightarrow \mu = k\left[ {\left( {9 + C} \right) - \left( {1 + C} \right)} \right]$
$ \Rightarrow \mu = k\left[ 8 \right]$
$ \Rightarrow \mu = \left( {0.45501} \right)\left[ 8 \right]$
$ \Rightarrow \mu = 3.64008$
Hence we get the required mean value
$\Rightarrow$${\sigma ^2} = \int\limits_{ - \infty }^\infty {{x^2}f\left( x \right)dx - {\mu ^2}} $
$\Rightarrow$$E\left( {{X^2}} \right) = \int\limits_{ - \infty }^\infty {{x^2}f\left( x \right)dx} $
Substitute the value of $f\left( x \right)$
$\Rightarrow$$E\left( {{X^2}} \right) = \int\limits_1^9 {{x^2}\dfrac{k}{x}dx} $
$\Rightarrow$$E\left( {{X^2}} \right) = \int\limits_1^9 {kxdx} $
$k$ is a constant and it can be taken outside
$\Rightarrow$$E\left( {{X^2}} \right) = k\int\limits_1^9 {xdx} $
$\Rightarrow$$E\left( {{X^2}} \right) = k\left[ {\dfrac{{{x^2}}}{2}} \right]_1^9$
Substitute the value of upper and lower limit in $x,$ that is upper limit minus lower limit, we get
$\Rightarrow$$E\left( {{X^2}} \right) = k\left[ {\left( {\dfrac{{{9^2}}}{2}} \right) - \dfrac{{{1^2}}}{2}} \right]$
Let us square the term and we get
$\Rightarrow$$E\left( {{X^2}} \right) = k\left[ {\left( {\dfrac{{81}}{2}} \right) - \dfrac{1}{2}} \right]$
On subtracting the term and we get
$\Rightarrow$$E\left( {{X^2}} \right) = k\left[ {\left( {\dfrac{{80}}{2}} \right)} \right]$
On rewriting the term and we get
$\Rightarrow$$E\left( {{X^2}} \right) = 0.45501\left[ {\left( {40} \right)} \right]$
Let us multiply the term and we get,
$\Rightarrow$$E\left( {{X^2}} \right) = 18.2004$
Now,
$\Rightarrow$${\sigma ^2} = Var\left( X \right) = E\left( {{X^2}} \right) - {\mu ^2}$
Let us putting the value and we get
$\Rightarrow$${\sigma ^2} = 18.2004 - {\left( {3.64008} \right)^2}$
On squaring the term and we get,
$\Rightarrow$${\sigma ^2} = 18.2004 - 13.2501$
Let us subtract the term and we get,
$\Rightarrow$${\sigma ^2} = 4.9503$
Hence we get the required variance.

Therefore the required mean and variance for the probability density functions are $3.64008$ and $4.9503$.

Note: The probability density function has the following properties:
$P\left( {a \leqslant x \leqslant b} \right) = \int_a^b {f\left( x \right)dx} $
It is non-negative for all real $X$
The probabilities are measured over intervals and not a single point.