Question

In each of the following, find the value of $k$ and find mean and variance of $X$ ;
I. P\left( x \right)=\left\\{ \begin{matrix} kx\ \ ,for\ x=1,2,3 \\\ 0\ \ \ \ ,otherwise \\\ \end{matrix} \right.

II.

$X=x$	-2	-1	0	1	2	3
$P\left( X=x \right)$	0.1	k	0.2	2k	0.3	K

Answer 1

Hint : In order to solve the given question, first we will find the value of ‘k’ using the formula $\sum {{P}_{i}}=1$ . Then by using the formula for calculating mean i.e. $E\left( X \right)=\mu =\sum\limits_{{}}^{{}}{i}=1n\times {{X}_{i}}{{P}_{i}}$ , we will find the value of mean of the given question. Later using the formula for calculating variance, first student need to find the value of $E\left( {{X}^{2}} \right)$ and ${{\left( E\left( X \right) \right)}^{2}}$ . Then subtracting the both will get you the required value of variance i.e. $Variance\left( X \right)=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}$ .
Formula used:
The mean of the random variable can also be said as the expectation of X.
The formula for calculating mean is given by,
$E\left( X \right)=\mu =\sum\limits_{{}}^{{}}{i}=1n\times {{X}_{i}}{{P}_{i}}$
$E\left( X \right)={{X}_{1}}{{P}_{1}}+{{X}_{2}}{{P}_{2}}+........+{{X}_{n}}{{P}_{n}}$
The formula for calculating the variance is given by:
$Variance\left( X \right)=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}$

Complete step-by-step answer :
I.

kx\ \ ,for\ x=1,2,3 \\\ 0\ \ \ \ ,otherwise \\\ \end{matrix} \right.$$ Here, It is given that, $$P\left( x \right)=kx\ \ ,for\ x=1,2,3$$ Thus, Putting the values of ‘x’, we will get $$\Rightarrow P\left( 1 \right)=k\ $$ $$\Rightarrow P\left( 2 \right)=2k\ $$ $$\Rightarrow P\left( 3 \right)=3k\ $$ As we know that, Sum of all the probabilities is equal to 1. Therefore, $$\Rightarrow k+2k+3k=1$$ $$\Rightarrow 6k=1$$ $$\Rightarrow k=\dfrac{1}{6}$$ Now, Finding the mean; The formula for mean is given by; $$E\left( X \right)=\sum\limits_{i=3}^{3}{{{X}_{i}}}P\left( {{X}_{i}} \right)$$ Substituting the values, we will get $$\Rightarrow E\left( X \right)=\left( \dfrac{1}{6}\left( 1 \right) \right)\left( 1 \right)+\left( \dfrac{1}{6}\left( 2 \right) \right)\left( 2 \right)+\left( \dfrac{1}{6}\left( 3 \right) \right)\left( 3 \right)$$ Solving the number, $$\Rightarrow E\left( X \right)=\dfrac{1}{6}+\dfrac{4}{6}+\dfrac{9}{6}=\dfrac{14}{6}$$ $$\Rightarrow E\left( X \right)=\dfrac{14}{6}$$ Now, Find the variance; The formula for calculating variance is given by; $$Variance\left( X \right)=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}$$ As we know that, $$E\left( X \right)=\sum\limits_{i=3}^{3}{{{X}_{i}}}P\left( {{X}_{i}} \right)=\dfrac{14}{6}$$ $${{\left( E\left( X \right) \right)}^{2}}={{\left( \dfrac{14}{6} \right)}^{2}}$$ Thus, $$E\left( {{X}^{2}} \right)=\sum\limits_{i=3}^{3}{{{X}_{i}}}P{{\left( {{X}_{i}} \right)}^{2}}$$ $$\Rightarrow E\left( {{X}^{2}} \right)=\left( \dfrac{1}{6}\left( 1 \right) \right){{\left( 1 \right)}^{2}}+\left( \dfrac{1}{6}\left( 2 \right) \right){{\left( 2 \right)}^{2}}+\left( \dfrac{1}{6}\left( 3 \right) \right){{\left( 3 \right)}^{2}}$$ Solving the number, $$\Rightarrow E{{\left( X \right)}^{2}}=\dfrac{1}{6}+\dfrac{8}{6}+\dfrac{27}{6}=\dfrac{36}{6}=6$$ $$\Rightarrow E{{\left( X \right)}^{2}}=6$$ Now, $$Variance\left( X \right)=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}$$ Substituting the values, $$Variance\left( X \right)=6-{{\left( \dfrac{14}{6} \right)}^{2}}=6-\dfrac{196}{36}=\dfrac{216}{36}-\dfrac{196}{36}=\dfrac{20}{36}=0.55$$ $$Variance\left( X \right)=0.55$$ Hence, this is the required answer. II. $$X=x$$ | -2| -1| 0| 1| 2| 3 ---|---|---|---|---|---|--- $$P\left( X=x \right)$$ | 0.1| k| 0.2| 2k| 0.3| K As we know that the probability distribution of the random variable is one. Therefore, $$\Rightarrow 0.1+k+0.2+2k+0.3+k=1$$ $$\Rightarrow 0.6+4k=1$$ $$\Rightarrow 4k=1-0.6$$ $$\Rightarrow k=\dfrac{0.04}{4}=\dfrac{4}{400}=\dfrac{1}{100}=0.01$$ $$\Rightarrow k=0.01$$ Therefore, $$X=x$$ | -2| -1| 0| 1| 2| 3 ---|---|---|---|---|---|--- $$P\left( X=x \right)$$ | 0.1| $$0.01$$ | 0.2| $$2k=2\left( 0.01 \right)=0.02$$ | 0.3| $$0.01$$ Now, Finding the mean; The formula for mean is given by; $$E\left( X \right)=\sum\limits_{{}}^{{}}{{{X}_{i}}}P\left( {{X}_{i}} \right)$$ Substituting the values, we will get $$\Rightarrow E\left( X \right)=0.1\left( -2 \right)+\left( 0.01 \right)\left( -1 \right)+\left( 0.2 \right)\left( 0 \right)+\left( 0.02 \right)\left( 1 \right)+\left( 0.3 \right)\left( 2 \right)+\left( 0.01 \right)\left( 3 \right)$$ Solving the number, $$\Rightarrow E\left( X \right)=-0.2-0.01+0+0.02+0.6+0.03$$ $$\Rightarrow E\left( X \right)=0.44$$ Now, Find the variance; The formula for calculating variance is given by; $$Variance\left( X \right)=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}$$ As we know that, $$E\left( X \right)=\sum\limits_{i=3}^{3}{{{X}_{i}}}P\left( {{X}_{i}} \right)=0.44$$ $${{\left( E\left( X \right) \right)}^{2}}={{\left( 0.44 \right)}^{2}}=0.1936$$ Thus, $$E\left( {{X}^{2}} \right)=\sum\limits_{{}}^{{}}{{{X}_{i}}}P{{\left( {{X}_{i}} \right)}^{2}}$$ $$\Rightarrow E\left( {{X}^{2}} \right)=0.1{{\left( -2 \right)}^{2}}+\left( 0.01 \right){{\left( -1 \right)}^{2}}+\left( 0.2 \right){{\left( 0 \right)}^{2}}+\left( 0.02 \right){{\left( 1 \right)}^{2}}+\left( 0.3 \right){{\left( 2 \right)}^{2}}+\left( 0.01 \right){{\left( 3 \right)}^{2}}$$ $$\Rightarrow E\left( {{X}^{2}} \right)=0.1\left( 4 \right)+\left( 0.01 \right)\left( 1 \right)+\left( 0.2 \right)\left( 0 \right)+\left( 0.02 \right)\left( 1 \right)+\left( 0.3 \right)\left( 4 \right)+\left( 0.01 \right)\left( 9 \right)$$ Solving the number, $$\Rightarrow E{{\left( X \right)}^{2}}=0.4+0.01+0+0.02+1.2+0.09$$ $$\Rightarrow E{{\left( X \right)}^{2}}=1.72$$ Now, $$Variance\left( X \right)=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}$$ Substituting the values, $$Variance\left( X \right)=1.72-0.1936$$ $$Variance\left( X \right)=1.526$$ Hence, this is the required answer. **Note** : In solving these types of questions, students should know the formula for calculating mean and variance of the given data. These types of examination are really marks maker and simple, students just note down the values from he given question or the problem very carefully to avoid making any type of error, else the answer can be wrong.