MCQ Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 4

The range of a random variable X is

${1, 2, 3, 4, ..}$ and the probabilities are

$P(X=K)=\dfrac{3^{CK}}{\angle K};$

$K=1,2,3,4$ ,......,then the value of C is

0%
$log_{e}3$
0%
$log_{e} 2$
0%
$log_{3}(log_{e}2)$
0%
$log_{2}(log_{e}3)$

Explanation

we know

${ e }^{ x }=\sum _{ 0 }^{ \infty }{ { x }^{ k }/k! } \\ and\quad \sum { P(i)=1\quad } \\ { e }^{ { 3 }^{ C } }={ e }^{ 0 }/0!+\sum _{ 1 }^{ \infty }{ P(x=k) } =1+1=2\\ { 3 }^{ C }\log _{ 2 }{ e } =\log _{ 2 }{ 2 } =1\\ C\quad =\quad \log _{ 3 }{ \log _{ e }{ 2 } }$

The range of a random variable

$\mathrm{X}$ is

$\left\{ 0,\quad 1,\quad 2 \right\}$ and

$\mathrm{P}(\mathrm{X}=0)=3\mathrm{K}^{3},\mathrm{P}(\mathrm{X}=1)=4\mathrm{K}-10\mathrm{K}^{2}$

$\mathrm{P}(\mathrm{x}=2)=5\mathrm{K}-1$ . Then we have

0%
$\mathrm{P}(\mathrm{X}=0)<\mathrm{P}(\mathrm{X}=2)<\mathrm{P}(\mathrm{X}=1)$
0%
$\mathrm{P}(\mathrm{X}=0)<\mathrm{P}(\mathrm{X}=1)<\mathrm{P}(\mathrm{X}=2)$
0%
$\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=2)$
0%
$\mathrm{P}(\mathrm{X}=1)>\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=2)$

Explanation

Using

$\sum P(X) = 1$

$\Rightarrow 3K^3-10K^2+4k+5K-1=3K^3-10K^2+9k-1=1\Rightarrow 3K^3-10K^2+9K-2=0$

Solving this we get

$K = \cfrac{1}{3},1,2$ but

$0<K<1$ so

$K=\dfrac{1}{3}$

Thus

$P(X=0) = \cfrac{1}{9}, P(X = 1) = \cfrac{2}{9}, P(X = 2) = \cfrac{2}{3}$

Hence order is

$\mathrm{P}(\mathrm{X}=0)<\mathrm{P}(\mathrm{X}=1)<\mathrm{P}(\mathrm{X}=2)$

Write the probability distribution when three coins are tossed.

0%
$X\quad \quad \quad :\begin{matrix} 0 & 1 & \quad 2 & 3 \end{matrix}\quad \\ P(X)\quad :\quad \begin{matrix} \cfrac { 1 }{ 8 } & \cfrac { 3 }{ 8 } & \cfrac { 3 }{ 8 } & \cfrac { 1 }{ 8 } \end{matrix}$
0%
$X\quad \quad \quad :\begin{matrix} 0 & 1 & \quad 2 & 3 \end{matrix}\quad \\ P(X)\quad :\quad \begin{matrix} \cfrac { 1 }{ 8 } & \cfrac { 3 }{ 8 } & \cfrac { 5 }{ 8 } & \cfrac { 7 }{ 8 } \end{matrix}$
0%
$X\quad \quad \quad :\begin{matrix} 0 & 1 & \quad 2 & 3 \end{matrix}\quad \\ P(X)\quad :\quad \begin{matrix} \cfrac { 7 }{ 8 } & \cfrac { 5 }{ 8 } & \cfrac { 3 }{ 8 } & \cfrac { 1 }{ 8 } \end{matrix}$
0%
$X\quad \quad \quad :\begin{matrix} 0 & 1 & \quad 2 & 3 \end{matrix}\quad \\ P(X)\quad :\quad \begin{matrix} \cfrac { 1 }{ 8 } & \cfrac { 3 }{ 8 } & \cfrac { 5 }{ 8 } & \cfrac { 1 }{ 8 } \end{matrix}$

Explanation

Assume

$X=$ number of heads for the coin.
Sample space

$=\left\{ HHH,HHT,HTT,TTT,HTH,THT,THH,TTH \right\}$
Hence,

$n(S)=8$
For no heads,we have

$n(X=0)=1\Rightarrow P(X=0)=\dfrac { 1 }{ 8 }$

For one heads,we have

$n(X=1)=3\Rightarrow P(X=1)=\dfrac { 3 }{ 8 }$

For two heads,we have

$n(X=2)=3\Rightarrow P(X=2)=\dfrac { 3 }{ 8 }$
similarly,for three heads,we get

$n(X=3)=1\Rightarrow P(X=3)=\dfrac { 1 }{ 8 }$

A player tosses two fair coins. He wins

$Rs.\ 5/-$ if two heads occur,

$Rs.$

$2/-$ if one head occurs and

$Rs.$

$1/-$ if no head occurs. Then his expected gain is

0%
$Rs.\dfrac{8}{3}$
0%
$Rs.\dfrac{7}{3}$
0%
$Rs.2.5$
0%
$Rs.1.5$

Explanation

Expected value

$=\sum { x(i)P(x(i)) }$
Expected gain

$= 5*1/4+2*1/2+1*1/4 = 2.5$

The probability of age of the workers to be 40 years or more is:

0%
$0.275$
0%
$0.475$
0%
$0.675$
0%
$0.975$

Explanation

Total no. of workers

$= 38+27+86+48+3=202$
Number of worker's age

$>40 = 86+48+3=137$

$P(E)= \cfrac {137}{202}=0.678$

$X$	1	2	3	4	5	6	7	8
$P(X)$	15	23	12	10	20	8	7	5

A random variable

$X$ has the probability distribution for the event

$\displaystyle E=\{ X$ is a prime number

$\}$ . If

$\displaystyle F=\left\{X<4\right\}$ , then the probability

$\displaystyle P(E\cup F)$ is

0%
$\displaystyle 0.40$
0%
$\displaystyle 0.77$
0%
$\displaystyle 0.57$
0%
$\displaystyle 0.45$

Explanation

From the given table prime numbers are

$2, 3, 5, 7$
'E' denotes prime number
'F' denotes the number

$\displaystyle <4$

$\displaystyle P(E)=P(2 \:or\: 3\: or\: 5\: or\: 7\:)$ ...(Events 2, 3, 5, 7 are in E)

$\displaystyle P(2)+P(3)+P(5)+P(7)=62$

$\displaystyle P(F)=P(1\:or 2 \: or 3)$ ...(Events 1,2, 3 are in E.)

$\displaystyle P(1)+P(2)+P(3)=50$

$\displaystyle P(E\cap F)=P(2 or 3)=P(2)+P(3)=35$

$\displaystyle P(E\cup F)=P(E)+P(F)-P(E\cap F)$

$\displaystyle =0.62+0.50-0.35 = 0.77$

Which set is shaded in the above diagram?

0%
$A\cap C$
0%
$A\cap B\cap C$
0%
$A\cup(B\cap C)$
0%
$A\cap(B\cup C)$

Explanation

$=A\cup(B \cap C)$

Hence, this is the answer.

Expectation of

$X$ equals

0%
$6$
0%
$12$
0%
$8$
0%
None of these

Explanation

Probability of getting 6 in one throw is,

$p =\cfrac{1}{6}$
Thus required expectation is,

$E(X)=\displaystyle \sum_{r=1}^{\infty}rP(X=r)=\sum_{r=1}^{\infty}r pq^{r-1} =\frac {p}{(1-q)^2}=\frac {p}{p^2}=\frac {1}{p}=6$

South African cricket captain lost the toss of a coin 13 times out ofThe chance of this happening was

0%
$\displaystyle \frac{7}{2^{13}}$
0%
$\displaystyle \frac{1}{2^{13}}$
0%
$\displaystyle \frac{13}{2^{14}}$
0%
$\displaystyle \frac{13}{2^{13}}$

Explanation

Tossing of a coin has only two outcomes which are equally likely hence it obviously follow binomial distribution

Let

$x$ be chances that he lost

$13$ times out of

$14$

where,

$p=$ probability=

$\dfrac 12$

$q=1-p=1-\dfrac 12=\dfrac 12,n=14$

Hence,

$p(X=x)=n(xp^x q^{n-x})$

$=^{14}C_{13}\left(\dfrac 12\right)^{13}\left(\dfrac 12\right)^{14-13}$

$p(X=x)=\dfrac {14!}{13!} \left(\dfrac 12\right)^{14}=14\times\left(\dfrac 12\right)^{14}=\dfrac 7{2^{13}}$

A fair die is tossed repeatedly until a six is obtained. Let

$X$ denote the number of tosses required. The probability that

$X\geq 3$ equals

0%
$\displaystyle \frac {125}{216}$
0%
$\displaystyle \frac {25}{36}$
0%
$\displaystyle \frac {5}{36}$
0%
$\displaystyle \frac {25}{216}$

Explanation

$P(X\leq 3)=1-[P(X=1)+P(X=2)]$

$=1-\left [\displaystyle \frac {1}{6}+\frac {1}{6}\left (\frac {5}{6}\right )\right ]$

$=1-\displaystyle \frac {11}{36}=\frac {25}{36}$

Let

$X$ represent the difference between the number of heads and the number of tails obtained when a coin is tossed

$6$ times. What are possible values of

$X$ ?

0%
$9,7,4,0$
0%
$0,2,4,6$
0%
$6,7,7,2$
0%
$6,4,2,0$

Explanation

A coin is tossed six times and

$X$ represents the difference between the number of heads and the number of tails.

$\therefore X\left( 6H,0T \right) =\left| 6-0 \right| =6$

$X\left( 5H,1T \right) =\left| 5-1 \right| =4$

$X\left( 4H,2T \right) =\left| 4-2 \right| =2$

$X\left( 3H,3T \right) =\left| 3-3 \right| =0$

$X\left( 2H,4T \right) =\left| 2-4 \right| =2$

$X\left( 1H,5T \right) =\left| 1-5 \right| =4$

$X\left( 0H,6T \right) =\left| 0-6 \right| =6$
Thus, the possible values of

$X$ are

$0, 2, 4$ and

$6$ .

The probability distribution of a random variable is given below:

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X = x)$	$0$	$K$	$2K$	$2K$	$3K$	$K^{2}$	$2K^{2}$	$7K^{2} + K$

Then

$P(0 < X < 5) =$

0%
$\dfrac {1}{10}$
0%
$\dfrac {3}{10}$
0%
$\dfrac {8}{10}$
0%
$\dfrac {7}{10}$

Explanation

$\sum P(X ) = 1$

$\Rightarrow 9K + 10K^{2} = 1$

$\Rightarrow 10K^{2} + 9K - 1 = 0$

$\Rightarrow 10K^{2} + 10K - K - 1 = 0$

$\Rightarrow 10K (K + 1) - 1(K + 1) = 0$

$\Rightarrow K = \dfrac {1}{10},-1$ (rejected)

$\therefore P(0 < X < 5) = P(X = 1) + P (X = 2) + P(X = 3) + P(X = 4)$

$= 8K$

$=\dfrac {8}{10}$

If p.d.f. of continuous r.v. is given by

$\displaystyle f\left( x \right) =\frac { x }{ 8 } ,0<x<4$

$\displaystyle =0$ , otherwise then F(x) is

0%
$\displaystyle \frac { { x }^{ 2 } }{ 8 }$
0%
$\displaystyle \frac { x }{ 4 }$
0%
$\displaystyle \frac { { x }^{ 2 } }{ 4 }$
0%
$\displaystyle \frac { { x }^{ 2 } }{ 16 }$

Explanation

$F(x) = \displaystyle\int \dfrac{x}{8} dx$

$\Rightarrow F(x) = \dfrac{x^2}{16}$

Let

$\displaystyle f\left( x \right) =\left\{ \begin{matrix} \dfrac { 1 }{ { x }^{ 2 } } ;1<x<\infty \\ 0;\text{otherwise} \end{matrix} \right\}$ , be the probability density function of random variable X, then the value of

$\displaystyle P\left( -2\le x\le 2 \right)$ is

0%
$2$
0%
$\log2$
0%
$\displaystyle \frac { 1 }{ 2 }$
0%
$0$

Explanation

Given

$\displaystyle f\left( x \right) =\left\{ \begin{matrix} \dfrac { 1 }{ { x }^{ 2 } } ;1<x<\infty \\ 0;\text{otherwise} \end{matrix} \right\}$

$P =\displaystyle \int_{-2}^{2} f(x) dx$

$\Rightarrow P = \displaystyle\int_{-2}^{1} 0 dx + \int_{1}^{2} \frac{1}{x^2} dx=0+\left[-\dfrac 1x\right]_{1}^{2}$

$\Rightarrow P = \displaystyle 0 + \left(\frac{-1}{2} + 1\right)$

$\Rightarrow P = \dfrac{1}{2}$

The mean of the numbers obtained on throwing a die having written 1 on three faces,

$2$ on two faces and

$5$ on one face is:

0%
$1$
0%
$2$
0%
$5$
0%
$\displaystyle\frac{8}{3}$

Explanation

$\textbf{Step-1: Assume the random variable to be equal to some variable & apply probability distribution. }$

$\text{Let}$

$X$

$\text{be the random variable representing a number on the die.}$

$\text{The total number of observations is six.}$

$\therefore P\left( X=1 \right) =\displaystyle\frac { 3 }{ 6 } =\displaystyle\frac { 1 }{ 2 }$

$P\left( X=2 \right) =\displaystyle\frac { 2 }{ 6 } =\displaystyle\frac { 1 }{ 3 }$

$P\left( X=5 \right) =\displaystyle\frac { 1 }{ 6 }$

$\text{Therefore, the probability distribution is as follows.}$

$\text{Mean}$

$E\left( X \right) =\sum { { p }_{ i }{ x }_{ i } }$

$=\displaystyle\frac { 1 }{ 2 } \times 1+\displaystyle\frac { 1 }{ 3 } \times 2+\displaystyle\frac { 1 }{ 6 } \times 5$

$\textbf{Step-2: Simplify the above results to get the required unknown.}$

$=\displaystyle\frac { 1 }{ 2 } +\displaystyle\frac { 2 }{ 3 } +\displaystyle\frac { 5 }{ 6 }$

$=\displaystyle\frac { 3+4+5 }{ 6 }$

$=\displaystyle\frac { 12 }{ 6 }$

$=2$

$\textbf{Hence, correct option is B}$

The probability distribution of

$x$ is

$x$	$0$	$1$	$2$	$3$
$P(x)$	$0.2$	$k$	$k$	$2k$

find the value of

$k$

0%
$0.2$
0%
$0.3$
0%
$0.4$
0%
$0.1$

Explanation

We must have,

$\sum P(x)=1$

$\Rightarrow 0.2+k+k+2k=1$

$\Rightarrow 4k=1-0.2=0.8$

$\therefore k=\dfrac{0.8}{4}=0.2$

Suppose that two cards are drawn at random from a deck of cards. Let

$X$ be the number of aces obtained. Then the value of

$E\left(X\right)$ is

0%
$\displaystyle\frac{37}{221}$
0%
$\displaystyle\frac{5}{13}$
0%
$\displaystyle\frac{1}{13}$
0%
$\displaystyle\frac{2}{13}$

Explanation

Let

$X$ denote the number of aces obtained. Therefore,

$X$ can take any of the values of 0, 1, or 2.
In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.

$\therefore P\left( X=0 \right) =P\left( 0 ace and 2 non-ace cards \right) =\displaystyle\frac { _{ }^{ 4 }{ { C }_{ 0 } }\times _{ }^{ 48 }{ { C }_{ 2 } } }{ _{ }^{ 52 }{ { C }_{ 2 } } } =\displaystyle\frac { 1128 }{ 1326 }$

$P\left( X=1 \right) =P\left( 1 ace and 1 non-ace cards \right) =\displaystyle\frac { _{ }^{ 4 }{ { C }_{ 1 } }\times _{ }^{ 48 }{ { C }_{ 1 } } }{ _{ }^{ 52 }{ { C }_{ 2 } } } =\displaystyle\frac { 192 }{ 1326 }$

$P\left( X=2 \right) =P\left( 2 ace and 0 non-ace cards \right) =\displaystyle\frac { _{ }^{ 4 }{ { C }_{ 2 } }\times _{ }^{ 48 }{ { C }_{ 0 } } }{ _{ }^{ 52 }{ { C }_{ 2 } } } =\displaystyle\frac { 6 }{ 1326 }$
Thus, the probability distribution is as follows.
Then,

$E\left( X \right) =\sum { { p }_{ i }{ x }_{ i } }$

$=0\times \displaystyle\frac { 1128 }{ 1326 } +1\times \displaystyle\frac { 192 }{ 1326 } +2\times \displaystyle\frac { 6 }{ 1326 }$

$=\displaystyle\frac { 204 }{ 1326 }$

$=\displaystyle\frac { 2 }{ 13 }$
Therefore, the correct answer is D.

For the probability distribution function of random variable X ,

$x_1,x_2,x_3,....x_n$ are the values X takes, and

$p(x_i)$ denote the probability of

$x_i$ then which one of the following is true.?

0%
$p(x_i)>0$
0%
$p(x_i)<0$
0%
$\sum p(x_i)=1$
0%
$\sum x_i=1$

Explanation

The probability distribution function of a random variable is a list of probabilities associated with each of its possible values.

Suppose a random variable

$X$ may take

$k$ different values, with the probability that

$X=x_i$ defined to be

$P(X=x_i)=p_i$ . The probabilities

$p_i$ must satify the following:

$0\leq p_i\leq1$ for each

$i$ .

$p_1+p_2+...+p_k=1$ that is

$\sum p_i=1$

Thus

$(A)$ and

$(C)$ are true.

The following is probabilty distribution of r.v X.

x	1	2	3	4	5	6
p(x)	$\frac{k}{6}$	$\frac{k}{6}$	$\frac{k}{6}$	$\frac{k}{6}$	$\frac{k}{6}$	$\frac{k}{6}$

then value of k is

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

We know that, Suppose a random variable

$X$ may take

$k$ different values, with the probability that

$X=x_i$ defined to be

$P(X=x_i)=p_i$ . The probabilities

$p_i$ must satify the following:

$0\leq p_i\leq1$ for each $i$ .
$p_1+p_2+...+p_k=1$ that is $\sum p_i=1$

Thus we have

$\sum p_i=1$

$\Rightarrow p_1+p_2+p_3+p_4+p_5+p_6=1$

$\Rightarrow \dfrac{k}{6}+\dfrac{k}{6}+\dfrac{k}{6}+\dfrac{k}{6}+\dfrac{k}{6}+\dfrac{k}{6}=1$

$\Rightarrow \dfrac{6k}{6}=1$

$\Rightarrow k=1$

Choose the discrete random varaiable of the following.

0%
Number of heads on a coin
0%
Sum of digits of uppermost face when two dice are rolled
0%
Height of students
0%
Weight of students

The variable which takes some specific values is called

0%
discrete random variable
0%
continuous random variable
0%
both A and B
0%
none

Explanation

A random variable is a variable whose value is a numerical outcome of a random experiment.

A discrete random variable

$X$ has a countable number of possible values.

That is, a random variable that takes some specific values is called discrete random variable.

Choose the contineous random variable of the following:

0%
Number of heads on a coin
0%
Sum of digits of uppermost face when two dice are rolled
0%
Height of students
0%
Weight of students

A biased coin is tossed twice.The probability of head is twice the tail.The PDF of number of heads is

x	$0$	$1$	$2$
p(x)	$\dfrac{a}{d}$	$\dfrac{b}{d}$	$\dfrac{c}{d}$

then values of

$a,b,c,d$ are

0%
$a=1,b=2,c=3,d=4$
0%
$a=1,b=4,c=4,d=9$
0%
$a=1,b=4,c=4,d=10$
0%
$a=1,b=2,c=1,d=4$

Explanation

A biased coin is tossed twice.

$\Rightarrow$ Probability of getting head,

$P(H)=\dfrac{2}{3}$

$\Rightarrow$ Probability of getting tail,

$P(T)=\dfrac{1}{3}$

Probability of getting head,

$1.$ PDF

$=$

Hence, the answer is

$a=1,b=4,c=4,d=9.$

The variable which assumes all possible values in the given interval is called

0%
discrete random variable
0%
continuous variable
0%
both A and B
0%
none

Which of the following is correct regarding probability mass function?

0%
All probabilities are positive.
0%
Any event in the distribution has a probability of happening of between 0 and 1.
0%
Both are correct
0%
None of the above

Explanation

Properties of probability mass function is listed below

$0\leq P_X(x) \leq 1$
$\sum_x P_X(x)=1$

Thus from the properties it is clear that, all probabilities are positive and any event in the distribution has a probability of happening of between

$0$ and

$1$ .

Hence, option C is correct.

The probability function is always

0%
Negative
0%
Non negative
0%
Positive
0%
None

Explanation

We know that, a PDF is valid if it satisfies the following

$(1)$

$f(x)$ is positive everywhere.

$(2)$ The area under the curve

$f(x)$ in the range

$(a,b)$ is 1, that is:

$\int _{a}^{b} f(x) dx=1$ .

Thus from

$(1)$ , we can say that the probability function is always non negative.

Standard normal probability distribution has mean equal to $40$ , whereas value of random variable x is $80$ and z-statistic is equal to $1.8$ then standard deviation of standard normal probability distribution is

0%
$150$
0%
$80$
0%
$40$
0%
$11.11$

Explanation

Given Mean

$\bar x=40$

$x=20$

$z=-1.8$

$\Rightarrow z=\dfrac { x-\bar { x } }{ \sigma }$

$\rightarrow \sigma= \dfrac { x-\bar { x }}{z}$

$\Rightarrow \sigma=\dfrac{20-40}{-1.8}=11.11$

Hence, the answer is

$11.11.$

The PDF of variable x: number of times sum 6 appears on in two throw of a pair of dice is

x	$0$	$1$	$2$
p(x)	$a$	$b$	$c$

then values of

$a,b,c$ are.

0%
$a=\dfrac{2}{36},b=\dfrac{3}{36},c=\dfrac{5}{36}$
0%
$a=\dfrac{961}{36},b=\dfrac{310}{36},c=\dfrac{25}{36}$
0%
$a=\dfrac{961}{1269},b=\dfrac{310}{1296},c=\dfrac{25}{1296}$
0%
$a=\dfrac{91}{36},b=\dfrac{30}{36},c=\dfrac{25}{36}$

Explanation

$x:$ no. of times sum

$6$ appears on in two throw of a pair of die.

$[(1,1)\;\;(1,2)\;\;(1,3)\;\;(1,4)\;\;(1,5)\;\;(1,6)$

$(2,1)\;\;(2,2)\;\;(2,3)\;\;(2,4)\;\;(2,5)\;\;(2,6)$

$(3,1)\;\;(3,2)\;\;(3,3)\;\;(3,4)\;\;(3,5)\;\;(3,6)$

$(4,1)\;\;(4,2)\;\;(4,3)\;\;(4,4)\;\;(4,5)\;\;(4,6)$

$=$ Sample Space

$(5,1)\;\;(5,2)\;\;(5,3)\;\;(5,4)\;\;(5,5)\;\;(5,6)$

$(6,1)\;\;(6,2)\;\;(6,3)\;\;(6,4)\;\;(6,5)\;\;(6,6)]$

$\Rightarrow P$ (getting sum six )

$=\dfrac{5}{36}$

$\Rightarrow P$ (getting sum other than six )

$=\dfrac{31}{36}$

Hence, the answer are :

$a=\dfrac{961}{1296}$

$b=\dfrac{310}{1296}$

$c=\dfrac{25}{1296}.$

From

$30$ bulbs out of which

$10$ are defective,

$3$ bulbs are chosen. X denotes number of defective bulbs.The PDF of r.v x is

x	0	1	2	3
p(x)	57k	95k	45k	6k

then k is

0%
$\dfrac{1}{210}$
0%
$\dfrac{2}{203}$
0%
$\dfrac{1}{203}$
0%
$\dfrac{1}{29}$

Explanation

We know that, Suppose a random variable

$X$ may take

$k$ different values, with the probability that

$X=x_i$ defined to be

$P(X=x_i)=p_i$ . The probabilities

$p_i$ must satify the following:

$0\leq p_i\leq1$ for each $i$ .
$p_1+p_2+...+p_k=1$ that is $\sum p_i=1$

Thus we have

$\sum p_i=1$

$\Rightarrow p_0+p_1+p_2+p_3=1$

$\Rightarrow 57k+95k+45k+6k=1$

$\Rightarrow 203k=1$

$\Rightarrow k=\dfrac{1}{203}$

The following is the p.d.f. (probability density function) of a continuous random variable

$X$ :

$f(x) = \dfrac {x}{32}, 0 < x < 8 = 0$ , otherwise
Find the expression for c.d.f. (cumulative distribution function) of

$X$

0%
$\cfrac {x^2}{64}-1$
0%
$\cfrac {x^2}{64}$
0%
$\cfrac {x+1}{64}$
0%
$\cfrac {x^2-1}{64}$

Explanation

c.d.f. of a continuous random variable

$X$ is given by,

$f(x) = \displaystyle \int_{-\infty}^{x} f(y) dy$

$\therefore f(x) = \displaystyle \int_{0}^{x}f(y) dy = \displaystyle \int_{0}^{x} \dfrac {y}{32}dy$

$= \left [\dfrac {y^{2}}{64}\right ]_{0}^{x} = \dfrac {x^{2}}{64}$

Thus,

$F(x) = \dfrac {x^{2}}{64}, x\in R$ .

CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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