Random Variable and Mathematical Expectation Exercise 6.2

Dear Class 12 Samacheer Kalvi students, here are the Text Book Solutions for Exercise 6.2 Business Maths Chapter 6 Random Variable and Mathematical for your reference and study.

You can find links to the other exercises below.

Important Formulas in Random Variable and Mathematical Expectation

Text Book Solutions for Differential Equations Exercise 6.1

Text Book Solutions for Differential Equations Exercise 6.3

Exercise 6.2

$\text{1. Find the expected value for the random variable of an unbiased die}\\$

$\text{Let X be the numbers on a die.} \\$
$X=1,2,3,4,5,6 \\$
$\text{The probability mass function is} \\$

X	1	2	3	4	5	6
p(X)	1/6	1/6	1/6	1/6	1/6	1/6

$E(X)=\sum_{}^{}x\;p(x) \\$
$=1\times\dfrac{1}{6}+2\times\dfrac{1}{6}+3\times\dfrac{1}{6} +4\times\dfrac{1}{6}+5\times\dfrac{1}{6}+6\times\dfrac{1}{6} \\$
$=\dfrac{1}{6}\left( 1+2+3+4+5+6 \right) \\$
$=\dfrac{21}{6} \\$
$=\dfrac{7}{2} \\$

$\text{2. Let X be a random variable defining number of students getting A }\\$
$\text{grade. Find the expected value of X from the given table} \\$

X=x	0	1	2	3
p(X=x)	0.2	0.1	0.4	0.3

$E(X)=\sum_{}^{}x\;p(x) \\$
$=0(0.2)+1(0.1)+2(0.4)+3(0.3) \\$
$=0+0.1+0.8+0.9 \\$
$=1.8 \\$

$\text{3. The following table is describing about the probability}\\$
$\text{ mass function of the random variable X} \\$

X=x	3	4	5
p(X=x)	0.2	0.3	0.5

$\text{Find the standard deviation of x.} \\$

$E(X)=\sum_{}^{}x\;p(x) \\$
$=3(0.1)+4(0.1)+5(0.2) \\$
$=0.3+.04+1 \\$
$E(x)=1.7 \\$

$E(X)^{2}=\sum_{}^{}x^{2}\;p(x) \\$
$=9(0.1)+16(0.1)+25(0.2) \\$
$=.9+1.6+5 \\$
$E(X)^{2}=7.5 \\$
$Var(X)=E(X)^{2}-\left[ E(x) \right]^{2} \\$
$=7.5-(1.7)^{2} \\$
$=7.5-2.89 \\$
$Var(X)=4.61 \\$
$S.D(x)=\sqrt{Var(X)} \\$
$=\sqrt{4.61} \\$
$S.D(x)=2.15 \\$

$\text{4. Let X be a continuous random variable with probability density function} \\$
$f_{x}(x)= \begin{cases} & 2x \hspace{5 mm} 0 \le x\le 1 \\ & 0 \hspace{8 mm} otherwise \end{cases}\\$
$\text{ Find the expected value of X} \\$
$E(X)=\displaystyle \int_{-\infty}^{\infty}x f(x)\;dx \\$
$=\displaystyle \int_{0}^{1}x (2x)\;dx \\$
$=2\displaystyle \int_{0}^{1}x^{2}\;dx \\$
$=2\bigg( \dfrac{x^{3}}{3} \bigg)_{0}^{1} \\$
$=2\bigg( \dfrac{1}{3}-0 \bigg) \\$
$E(X)=\dfrac{2}{3} \\$

$5. \text{Let X be a continuous random variable with probability density function} \\$
$f(x)= \begin{cases} & \dfrac{3}{x^{4}} \hspace{5 mm} x\ge 1 \\ & 0 \hspace{8 mm} otherwise \end{cases} \\$
$\text{Find the mean and variance of X.} \\$

$E(x)=\displaystyle \int_{-\infty}^{\infty}x f(x)dx \\$
$=\displaystyle \int_{1}^{\infty}x \left( \dfrac{3}{x^{4}} \right)\;dx \\$
$=3\displaystyle \int_{1}^{\infty}\dfrac{1}{x^{3}}\;dx \\$
$=3\displaystyle \int_{1}^{\infty}x^{-3}\;dx \\$
$=3\bigg( \dfrac{x^{-2}}{-2} \bigg)_{1}^{\infty} \\$
$=\dfrac{-3}{2}\bigg( \dfrac{1}{x^{2}} \bigg)_{1}^{\infty} \\$
$=\dfrac{-3}{2}\left( \dfrac{1}{\infty}-1 \right) \\$
$=\dfrac{-3}{2} (0-1) \\$
$E(x)=\dfrac{3}{2} \\$

$E(x^{2})=\displaystyle \int_{-\infty}^{\infty}x^{2} f(x)dx \\$
$=\displaystyle \int_{1}^{\infty}x^{2} \left( \dfrac{3}{x^{4}} \right)\;dx \\$
$=3\displaystyle \int_{1}^{\infty}\dfrac{1}{x^{2}}\;dx \\$
$=3\displaystyle \int_{1}^{\infty}x^{-2}\;dx \\$
$=-3\bigg( \dfrac{1}{x} \bigg)_{1}^{\infty} \\$
$=-3\bigg(\dfrac{1}{\infty}-1\bigg) \\$
$=-3(0-1) \\$
$E(x^{2})=3 \\$
$Mean=E(x)=\dfrac{3}{2} \\$
$Var(x)=E(x^{2})-\left[ E(x) \right]^{2} \\$
$=3-\left( \dfrac{3}{2} \right)^{2} \\$
$=3-\dfrac{9}{4} \\$
$=\dfrac{12-9}{4} \\$
$Var (x)=\dfrac{3}{4} \\$

$\text{6.In an investment, a man can make a profit of Rs. 5,000 with a}\\$
$\text{probability of 0.62 or a loss of Rs. 8,000 with a probability of.}\; 0.38. \\$
$\text{ Find the expected gain.} \\$

$\text{Let X be the amount} \\$
$X=5000, -8000 \\$
$\text{The probability mass function is}\\$

x	5000	-8000
p(x)	0.62	0.38

$E(X)=\sum_{}^{}x p(x) \\$
$=5000(0.62)+(-8000)(0.38) \\$
$=3100-3040 \\$
$E(X)=60 \\$

$\text{7. What are the properties of Mathematical expectation?}\\$

$\text{(i)E(a)=a where 'a' is a constant} \\$
$\text{(ii) E(aX)=aE(X)} \\$
$\text{(iii) E(aX+b)=aE(X)+b , where 'a' and 'b' are constants} \\$
$(iv) If X \ge 0, \;then \; E(X) \ge 0 \\$
$\text{(v) V (a)= 0} \\$
$\text{(vi) If X is random variable, then}\; V(aX+b)=a^{2 }V(X) \\$

$\text{8. What do you understand by Mathematical expectation?}\\$

$\text{Mathematical expectation (also called expected value) is the average}\\$
$\text{or mean value of a random variable that you would expect if an }\\$
$\text{experiment were repeated a large number of times.}\\$

$\text{ In simple terms, it represents the long-term average outcome of a random process.} \\$

$\text{9. How do you define variance in terms of Mathematical expectation?}\\$

$\text{Variance is a measure of how much the values of a random variable deviate} \\$
$\text{from its mean (expected value). In terms of mathematical expectation,} \\$
$\text{the variance of a random variable X is defined as:} \\$
$Var(X)=E(X)^{2}- \left[ E(X) \right]^{2} \\$

$\text{10. Define Mathematical expectation in terms of discrete random}\\$
$\text{ variable.}\\$

$\text{For a discrete random variable, the mathematical expectation (or expected value)} \\$
$\text{is defined as the weighted average of all possible values that the random variable} \\$
$\text{can take, where each value is weighted by its corresponding probability.} \\$
$\text{If a discrete random variable X takes values} \; x_{1}, x_{2},x_{3}... x_{n} \\$ $\;\text{with probabilities} \; p_{1}, p_{2},p_{3}...p_{n} \; \text{then its mathematical expectation is given by:}\\$
$E(X)=\sum_{i=1}^{n}x_{i}p_{i}\\$

$\text{11. State the definition of Mathematical expectation using} \\$ $\text{ continuous random variable.}\\$

$\text{For a continuous random variable, the mathematical expectation (or expected value)}\\$
$\text{is defined as the integral of the product of the variable and its probability}\\$
$\text{ density function over all \text{}possible values.}\\$
$\text{If X is a continuous random variable with probability density function }\\$
$\text{ (p.d.f.) f(x), then the mathematical expectation of X is given by:}\\$
$E(X)=\int_{-\infty}^{\infty} x f(x) dx\\$

$\text{12. In a business venture a man can make a profit of Rs. 2,000 with a}\\$
$\text{probability of 0.4 or have a loss of Rs.1,000 with a probability}\\$
$\text{of 0⋅6 . What is his expected, variance and standard deviation of profit?}\\$

$\text{Let X be the amount} \\$
$X=2000,-1000 \\$
$\text{The probability mass function is} \\$

x	2000	-1000
p(x)	.4	.6

$E(X)=\sum_{}^{}x p(x) \\$
$=2000(0.4)+(-1000)(0.6) \\$
$=800-600 \\$
$E(X)=200 \\$
$E(X)^{2}=\sum_{}^{}x^{2} p(x) \\$
$=4000000(0.4)+(1000000)(0.6) \\$
$=1600000+600000 \\$
$E(X)^{2}=2200000 \\$
$Var (X)= E(X)^{2}-\left[ E(X) \right]^{2}\\$
$=2200000-(200)^{2}\\$
$=2200000-40000 \\$
$Var(X)= 2160000\\$
$S.D (X)=\sqrt{Var(X)} \\$
$=\sqrt{216000} \\$
$=1469.69\\$

$\text{13. The number of miles an automobile tire lasts before it reaches a }\\$
$\text{critical point in tread wear can be represented by a p.d.f.}\\$
$f(x)= \begin{cases} & \dfrac{1}{30} e^{\frac{-x}{30}}\hspace{5 mm} for \;x>0 \\ & 0 \hspace{15 mm} for\;x \le0 \end{cases}\\$
$\text{Find the expected number of miles (in thousands) a tire would last until}\\$
$\text{it reaches the critical tread wear point.}\\$

$E(X)=\displaystyle \sum_{-\infty}^{\infty}x f(x) \\$
$=\displaystyle \sum_{0}^{\infty}x \left( \frac{1}{30} \right)e^{\frac{-x}{30}}dx \\$
$=\frac{1}{30}\sum_{0}^{\infty}x e^{\frac{-x}{30}}dx \\$
$a=\frac{1}{30}\; n=1\\$
$\frac{1}{30} \frac{1!}{\left( \frac{1}{30} \right)^{1+1}} \; \\$
$Since\bigg[\int_{0}^{\infty} x^{n}e^{-ax}dx=\frac{n!}{a^{n+1}} \bigg]\\$
$\frac{1}{30} \frac{1}{\left( \frac{1}{30} \right)^{2}} \\$
$=\frac{1}{\frac{1}{30}}\\$
$=E(X)=30\\$
$\text{No. of miles expected} = 30\times 1000\\$
$=30000 \;miles.\\$

$\text{14. A person tosses a coin and is to receive Rs. 4 for a head and is to} \\$
$\text{pay Rs. 2 for a tail. Find the expectation and variance of his gains.}\\$

$\text{Let X be the amount} \\$
$X=4,-2\\$
$\text{ The probability mass function is} \\$

x	4	-2
p(x)	1/2	1/2

$E(X)=\sum_{}^{}x p(x) \\$
$=4\times \frac{1}{2}-2 \times \frac{1}{2} \\$
$=2-1 \\$
$=1 \\$
$E(X)^{2}=\sum_{}^{}x^{2} p(x) \\$
$=16\times \frac{1}{2}+4\times \frac{1}{2} \\$
$=8+2\\$
$E(X)^{2}=10\\$
$Var(X)=E(X)^{2}-\left[ E(X) \right]^{2}\\$
$=10-(1)^{2} \\$
$=10-1\\$
$Var(X)=9\\$