Probability distribution Questions in English

(C) The sum of probabilities in a probability distribution is $1$.
$0.07 + 0.2 + 0.3 + k + 0.07 + 0.04 + 0.02 = 1$
$0.7 + k = 1 \implies k = 0.3$
Now,we calculate the mean $E(X) = \sum x_i p_i$ and $E(X^2) = \sum x_i^2 p_i$.

$x_i$	$p_i$	$x_i p_i$	$x_i^2 p_i$
$5$	$0.07$	$0.35$	$1.75$
$6$	$0.2$	$1.2$	$7.2$
$7$	$0.3$	$2.1$	$14.7$
$8$	$0.3$	$2.4$	$19.2$
$9$	$0.07$	$0.63$	$5.67$
$10$	$0.04$	$0.4$	$4$
$11$	$0.02$	$0.22$	$2.42$
Total	$1$	$7.3$	$55.04$

Mean $E(X) = \sum x_i p_i = 7.3$
$E(X^2) = \sum x_i^2 p_i = 55.04$
Variance $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$
$\operatorname{Var}(X) = 55.04 - (7.3)^2 = 55.04 - 53.29 = 1.75$

(D) die has $6$ faces. The numbers on the faces are $1, 1, 1, 2, 2, 5$.
The probability of getting $1$ is $P(X=1) = \frac{3}{6} = \frac{1}{2}$.
The probability of getting $2$ is $P(X=2) = \frac{2}{6} = \frac{1}{3}$.
The probability of getting $5$ is $P(X=5) = \frac{1}{6}$.
The mean (expected value) is given by $E(X) = \sum p_i x_i$.
Mean $= (1 \times \frac{1}{2}) + (2 \times \frac{1}{3}) + (5 \times \frac{1}{6}) = \frac{1}{2} + \frac{2}{3} + \frac{5}{6} = \frac{3+4+5}{6} = \frac{12}{6} = 2$.

$x_i$	$p_i x_i$
$1$	$1/2$
$2$	$2/3$
$5$	$5/6$
Total	$2$

(A) Since $f(x)$ is the p.d.f. of a random variable $X$,the total area under the curve must be $1$.
$\int_{0}^{1} f(x) dx = 1 \Rightarrow \int_{0}^{1} K(x-x^2) dx = 1$
$K \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = 1 \Rightarrow K \left( \frac{1}{2} - \frac{1}{3} \right) = 1$
$K \left( \frac{1}{6} \right) = 1 \Rightarrow K = 6$
Now,we calculate $P(X < \frac{1}{2}) = \int_{0}^{\frac{1}{2}} 6(x-x^2) dx$
$= 6 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{\frac{1}{2}} = \left[ 3x^2 - 2x^3 \right]_{0}^{\frac{1}{2}}$
$= 3 \left( \frac{1}{2} \right)^2 - 2 \left( \frac{1}{2} \right)^3 = 3 \left( \frac{1}{4} \right) - 2 \left( \frac{1}{8} \right)$
$= \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

(A) The mean of the random variable $X$ is given by $E(X) = \sum p_i x_i = \frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \dots + \frac{n}{n}$.
$E(X) = \frac{1}{n} (1 + 2 + 3 + \dots + n) = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$.
The expected value of $X^2$ is given by $E(X^2) = \sum p_i x_i^2 = \frac{1^2}{n} + \frac{2^2}{n} + \frac{3^2}{n} + \dots + \frac{n^2}{n}$.
$E(X^2) = \frac{1}{n} (1^2 + 2^2 + 3^2 + \dots + n^2) = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$.
The variance of $X$ is given by $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$.
$\operatorname{Var}(X) = \frac{(n+1)(2n+1)}{6} - \left( \frac{n+1}{2} \right)^2$.
$\operatorname{Var}(X) = \frac{2n^2 + 3n + 1}{6} - \frac{n^2 + 2n + 1}{4}$.
Taking the least common multiple as $12$:
$\operatorname{Var}(X) = \frac{2(2n^2 + 3n + 1) - 3(n^2 + 2n + 1)}{12} = \frac{4n^2 + 6n + 2 - 3n^2 - 6n - 3}{12} = \frac{n^2 - 1}{12}$.

(B) Given the probability mass function $P(X=x) = \frac{{}^{5}C_{x}}{2^{5}}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.
We need to calculate $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=0) = \frac{{}^{5}C_{0}}{2^{5}} = \frac{1}{32}$
$P(X=1) = \frac{{}^{5}C_{1}}{2^{5}} = \frac{5}{32}$
$P(X=2) = \frac{{}^{5}C_{2}}{2^{5}} = \frac{10}{32}$
Summing these values: $P(X \leq 2) = \frac{1+5+10}{32} = \frac{16}{32} = \frac{1}{2}$.
Now,check the options:
$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) = \frac{{}^{5}C_{3} + {}^{5}C_{4} + {}^{5}C_{5}}{2^{5}} = \frac{10+5+1}{32} = \frac{16}{32} = \frac{1}{2}$.
Since $P(X \leq 2) = \frac{1}{2}$ and $P(X \geq 3) = \frac{1}{2}$,we have $P(X \leq 2) = P(X \geq 3)$.

(A) For a probability density function,the total area under the curve must be $1$.
$\int_{0}^{4} f(x) dx = 1$
$\int_{0}^{4} \frac{k}{\sqrt{x}} dx = 1$
$k [2\sqrt{x}]_{0}^{4} = 1$
$k [2(2) - 0] = 1 \Rightarrow 4k = 1 \Rightarrow k = \frac{1}{4}$
Now,we calculate $P(1 < X < 4)$:
$P(1 < X < 4) = \int_{1}^{4} \frac{1/4}{\sqrt{x}} dx$
$= \frac{1}{4} [2\sqrt{x}]_{1}^{4}$
$= \frac{1}{2} [\sqrt{4} - \sqrt{1}]$
$= \frac{1}{2} [2 - 1] = \frac{1}{2}$

(D) For a probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$k + 0 + 2k + 5k + k + 3k = 1$.
$12k = 1 \Rightarrow k = \frac{1}{12}$.
We need to find $P(X \geq 4)$,which is $P(X = 4) + P(X = 5) + P(X = 6)$.
$P(X \geq 4) = 5k + k + 3k = 9k$.
Substituting the value of $k$,we get $P(X \geq 4) = 9 \times \frac{1}{12} = \frac{9}{12} = \frac{3}{4}$.

(D) For a probability mass function,the sum of all probabilities must be equal to $1$.
Therefore,$P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 1$.
Substituting the given values: $k + \frac{k}{3} + \frac{k}{4} + \frac{k}{2} + \frac{k}{2} = 1$.
Taking the least common multiple $(LCM)$ of the denominators $(3, 4, 2, 2)$,which is $12$:
$\frac{12k + 4k + 3k + 6k + 6k}{12} = 1$.
Summing the numerators: $\frac{31k}{12} = 1$.
Solving for $k$: $k = \frac{12}{31}$.

(C) The cumulative distribution function $F(x)$ is defined as $F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) dt$.
For $-1 < x < 2$,we have:
$F(x) = \int_{-1}^{x} \frac{t^3}{3} dt$
$F(x) = \frac{1}{3} \left[ \frac{t^4}{4} \right]_{-1}^{x}$
$F(x) = \frac{1}{3} \left( \frac{x^4}{4} - \frac{(-1)^4}{4} \right)$
$F(x) = \frac{1}{3} \left( \frac{x^4}{4} - \frac{1}{4} \right)$
$F(x) = \frac{1}{12} (x^4 - 1)$

(A) Given the p.d.f. $f(x) = \frac{1}{2}$ for $0 < x < 2$.
First,we calculate $a = P(X < \frac{1}{2})$:
$a = \int_{0}^{1/2} \frac{1}{2} dx = \frac{1}{2} [x]_{0}^{1/2} = \frac{1}{2} (\frac{1}{2} - 0) = \frac{1}{4}$.
Next,we calculate $b = P(X > \frac{3}{2})$:
$b = \int_{3/2}^{2} \frac{1}{2} dx = \frac{1}{2} [x]_{3/2}^{2} = \frac{1}{2} (2 - \frac{3}{2}) = \frac{1}{2} (\frac{1}{2}) = \frac{1}{4}$.
Comparing $a$ and $b$,we have $a = \frac{1}{4}$ and $b = \frac{1}{4}$.
Therefore,$a - b = \frac{1}{4} - \frac{1}{4} = 0$.

(C) The cumulative distribution function (c.d.f.) is given by $F(x) = P(X \leq x) = \frac{\sqrt{x}}{2}$.
We need to find the probability $P[X > 1]$.
Using the property of the cumulative distribution function,we know that $P(X > x) = 1 - P(X \leq x) = 1 - F(x)$.
Therefore,$P[X > 1] = 1 - F(1)$.
Substituting $x = 1$ into the given function: $F(1) = \frac{\sqrt{1}}{2} = \frac{1}{2}$.
Thus,$P[X > 1] = 1 - \frac{1}{2} = \frac{1}{2}$.

(B) For a probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$P(0) + P(1) + P(2) + P(3) + P(4) = 1$.
Substituting the given values: $k + 2k + 4k + 2k + k = 1$.
$10k = 1 \Rightarrow k = \frac{1}{10}$.
We need to find $P(X \leq 2)$,which is $P(0) + P(1) + P(2)$.
$P(X \leq 2) = k + 2k + 4k = 7k$.
Substituting the value of $k$: $7 \times \frac{1}{10} = \frac{7}{10}$.

(D) Since $f(x)$ is a probability density function,the total area under the curve must be $1$:
$\int_{-2}^{2} k(4 - x^2) dx = 1$
Since the function is even,we have $2k \int_{0}^{2} (4 - x^2) dx = 1$.
$2k [4x - \frac{x^3}{3}]_0^2 = 1$
$2k (8 - \frac{8}{3}) = 1 \Rightarrow 2k(\frac{16}{3}) = 1 \Rightarrow k = \frac{3}{32}$.
Now,we calculate $P[-1 < X < 1]$:
$P[-1 < X < 1] = \int_{-1}^{1} \frac{3}{32}(4 - x^2) dx = 2 \times \frac{3}{32} \int_{0}^{1} (4 - x^2) dx$
$= \frac{6}{32} [4x - \frac{x^3}{3}]_0^1 = \frac{3}{16} (4 - \frac{1}{3}) = \frac{3}{16} \times \frac{11}{3} = \frac{11}{16}$.

(B) The probability $P(|X| < 2)$ is equivalent to $P(-2 < X < 2)$.
Given the p.d.f. $f(x) = \frac{x+2}{18}$ for $-2 < x < 4$,we integrate over the interval $(-2, 2)$:
$P(-2 < X < 2) = \int_{-2}^{2} \frac{x+2}{18} dx$
$= \frac{1}{18} \left[ \frac{x^2}{2} + 2x \right]_{-2}^{2}$
$= \frac{1}{18} \left[ (\frac{2^2}{2} + 2(2)) - (\frac{(-2)^2}{2} + 2(-2)) \right]$
$= \frac{1}{18} \left[ (2 + 4) - (2 - 4) \right]$
$= \frac{1}{18} [6 - (-2)]$
$= \frac{8}{18} = \frac{4}{9}$

(C) For a probability mass function (p.m.f.),the sum of all probabilities must be equal to $1$,i.e.,$\sum P(X) = 1$.
Given $P(X) = k \binom{4}{x}$ for $x = 0, 1, 2, 3, 4$.
Calculating each probability:
$P(X=0) = k \binom{4}{0} = k \times 1 = k$
$P(X=1) = k \binom{4}{1} = k \times 4 = 4k$
$P(X=2) = k \binom{4}{2} = k \times 6 = 6k$
$P(X=3) = k \binom{4}{3} = k \times 4 = 4k$
$P(X=4) = k \binom{4}{4} = k \times 1 = k$
Summing these probabilities:
$k + 4k + 6k + 4k + k = 1$
$16k = 1$
$k = \frac{1}{16}$

(D) The mean $E(X)$ is calculated as:
$E(X) = \sum x_i P_i = (-2)(0.2) + (-1)(0.3) + (0)(0.1) + (1)(0.15) + (2)(0.25)$
$E(X) = -0.4 - 0.3 + 0 + 0.15 + 0.5 = -0.05$
Next,we calculate $E(X^2)$:
$E(X^2) = \sum x_i^2 P_i = (-2)^2(0.2) + (-1)^2(0.3) + (0)^2(0.1) + (1)^2(0.15) + (2)^2(0.25)$
$E(X^2) = (4)(0.2) + (1)(0.3) + 0 + (1)(0.15) + (4)(0.25)$
$E(X^2) = 0.8 + 0.3 + 0.15 + 1.0 = 2.25$
The variance $Var(X)$ is given by:
$Var(X) = E(X^2) - [E(X)]^2$
$Var(X) = 2.25 - (-0.05)^2$
$Var(X) = 2.25 - 0.0025 = 2.2475$

(C) The given probability mass function is $P(X=x) = \frac{\binom{5}{x}}{2^5}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.
Calculating the probabilities:
$P(X=0) = \frac{\binom{5}{0}}{32} = \frac{1}{32}$,$P(X=1) = \frac{\binom{5}{1}}{32} = \frac{5}{32}$,$P(X=2) = \frac{\binom{5}{2}}{32} = \frac{10}{32}$,$P(X=3) = \frac{\binom{5}{3}}{32} = \frac{10}{32}$,$P(X=4) = \frac{\binom{5}{4}}{32} = \frac{5}{32}$,$P(X=5) = \frac{\binom{5}{5}}{32} = \frac{1}{32}$.
Now evaluate the options:
$A$. $P(X \leq 1) = P(0) + P(1) = \frac{1+5}{32} = \frac{6}{32}$ and $P(X \geq 4) = P(4) + P(5) = \frac{5+1}{32} = \frac{6}{32}$. Thus,$P(X \leq 1) = P(X \geq 4)$ is true.
$B$. $P(X \leq 2) = P(0) + P(1) + P(2) = \frac{1+5+10}{32} = \frac{16}{32} = 0.5$. $P(X \geq 4) = \frac{6}{32} = 0.1875$. Since $0.5 \geq 0.1875$,this is true.
$C$. $P(X \leq 3) = P(0) + P(1) + P(2) + P(3) = \frac{1+5+10+10}{32} = \frac{26}{32} = 0.8125$. $P(X \geq 3) = P(3) + P(4) + P(5) = \frac{10+5+1}{32} = \frac{16}{32} = 0.5$. Since $0.8125 \leq 0.5$ is false,option $C$ is not true.
$D$. $P(X \leq 2) = \frac{16}{32} = 0.5$ and $P(X \geq 3) = \frac{16}{32} = 0.5$. Thus,$P(X \leq 2) = P(X \geq 3)$ is true.

(C) We are given the probability density function $f(x) = \frac{x+2}{18}$ for $-2 < x < 4$.
We need to find $P[|x| < 1]$.
The condition $|x| < 1$ is equivalent to $-1 < x < 1$.
Thus,$P[|x| < 1] = \int_{-1}^{1} f(x) \, dx = \int_{-1}^{1} \frac{x+2}{18} \, dx$.
Evaluating the integral:
$P = \frac{1}{18} \int_{-1}^{1} (x+2) \, dx = \frac{1}{18} \left[ \frac{x^2}{2} + 2x \right]_{-1}^{1}$.
Substituting the limits:
$P = \frac{1}{18} \left[ (\frac{1}{2} + 2) - (\frac{(-1)^2}{2} + 2(-1)) \right]$.
$P = \frac{1}{18} \left[ (\frac{1}{2} + 2) - (\frac{1}{2} - 2) \right]$.
$P = \frac{1}{18} [ \frac{1}{2} + 2 - \frac{1}{2} + 2 ] = \frac{1}{18} [4] = \frac{4}{18} = \frac{2}{9}$.
Therefore,the correct option is $C$.

(B) When a die is thrown,the possible outcomes are $S = \{1, 2, 3, 4, 5, 6\}$.
The probability of getting any number $x \in S$ is $P(x) = \frac{1}{6}$.
The expectation $E(X)$ is given by the formula $E(X) = \sum x \cdot P(x)$.
$E(X) = (1 \times \frac{1}{6}) + (2 \times \frac{1}{6}) + (3 \times \frac{1}{6}) + (4 \times \frac{1}{6}) + (5 \times \frac{1}{6}) + (6 \times \frac{1}{6})$.
$E(X) = \frac{1}{6} \times (1 + 2 + 3 + 4 + 5 + 6)$.
$E(X) = \frac{1}{6} \times 21 = 3.5$.

(C) To find $P(X \leq 2)$,we integrate the probability density function $f(x)$ from the lower bound $0$ to $2$.
$P(X \leq 2) = \int_{0}^{2} f(x) \, dx$
$P(X \leq 2) = \int_{0}^{2} \frac{x}{8} \, dx$
$P(X \leq 2) = \frac{1}{8} \left[ \frac{x^2}{2} \right]_{0}^{2}$
$P(X \leq 2) = \frac{1}{8} \left( \frac{2^2}{2} - \frac{0^2}{2} \right)$
$P(X \leq 2) = \frac{1}{8} \left( \frac{4}{2} \right) = \frac{1}{8} \times 2 = \frac{2}{8} = \frac{1}{4}$

(B) To find the standard deviation,we first calculate the mean $\mu = E(X)$ and the variance $\sigma^2 = E(X^2) - [E(X)]^2$.
The probability distribution is:

$x_i$	$p_i$	$p_i x_i$	$p_i x_i^2$
$0$	$q^2$	$0$	$0$
$1$	$2pq$	$2pq$	$2pq$
$2$	$p^2$	$2p^2$	$4p^2$
Total	$1$	$2pq + 2p^2$	$2pq + 4p^2$

Mean $\mu = E(X) = \sum p_i x_i = 2pq + 2p^2 = 2p(q+p)$.
Since $p+q=1$,we have $\mu = 2p(1) = 2p$.
Variance $\sigma^2 = E(X^2) - \mu^2 = (2pq + 4p^2) - (2p)^2 = 2pq + 4p^2 - 4p^2 = 2pq$.
Standard deviation $\sigma = \sqrt{\text{Variance}} = \sqrt{2pq}$.

(B) For a discrete random variable $X$,the probability mass function $P[X=x]$ is related to the cumulative distribution function $F(x)$ by the formula:
$P[X=x] = F(x) - F(x^-)$
where $F(x^-)$ is the value of the cumulative distribution function at the value immediately preceding $x$.
In this case,we want to find $P[X=3]$.
Looking at the table,the value of $X$ immediately preceding $3$ is $1$.
Therefore,$P[X=3] = F(3) - F(1)$.
From the table:
$F(3) = 0.75$
$F(1) = 0.65$
Substituting these values:
$P[X=3] = 0.75 - 0.65 = 0.10$
Thus,the correct option is $(B)$.

(B) The cumulative distribution function $F(x)$ is defined as $F(x) = P(X \leq x)$.
We need to calculate $F(1.5) - F(-0.5) = P(X \leq 1.5) - P(X \leq -0.5)$.
By the definition of probability for a discrete random variable,$P(X \leq x) = \sum_{x_i \leq x} P(X = x_i)$.
Therefore,$P(X \leq 1.5) = P(X = -1.5) + P(X = -0.5) + P(X = 0.5) + P(X = 1.5) = 0.05 + 0.2 + 0.15 + 0.25 = 0.65$.
And $P(X \leq -0.5) = P(X = -1.5) + P(X = -0.5) = 0.05 + 0.2 = 0.25$.
Thus,$F(1.5) - F(-0.5) = 0.65 - 0.25 = 0.4$.
Alternatively,$F(1.5) - F(-0.5) = P(-0.5 < X \leq 1.5) = P(X = 0.5) + P(X = 1.5) = 0.15 + 0.25 = 0.4$.
The correct option is $B$.

(A) The cumulative distribution function $F(x)$ is defined as the integral of the probability density function $f(x)$.
$F(x) = \int_{0}^{x} f(t) dt = \int_{0}^{x} 3(1 - 2t^2) dt$.
Evaluating the integral:
$F(x) = 3 \left[ t - \frac{2t^3}{3} \right]_{0}^{x} = 3 \left( x - \frac{2x^3}{3} \right)$.
Comparing this with the given form $F(x) = k(x - \frac{2x^3}{k})$,we observe that $k = 3$ is not directly matching the structure unless we rewrite the expression.
Wait,let us re-evaluate the expression: $F(x) = 3x - 2x^3$.
If $F(x) = k(x - \frac{2x^3}{k}) = kx - 2x^3$,then by comparing coefficients,$k = 3$ and $2 = 2$. Thus,$k = 3$.

(A) For a probability distribution,the sum of all probabilities must be $1$.
Therefore,$k + 3k + 3k + k = 1$,which implies $8k = 1$,so $k = \frac{1}{8}$.
The mean $E(X) = \sum x_i P(x_i) = (0 \times \frac{1}{8}) + (1 \times \frac{3}{8}) + (2 \times \frac{3}{8}) + (3 \times \frac{1}{8}) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2}$.
The expected value of the square,$E(X^2) = \sum x_i^2 P(x_i) = (0^2 \times \frac{1}{8}) + (1^2 \times \frac{3}{8}) + (2^2 \times \frac{3}{8}) + (3^2 \times \frac{1}{8}) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$.
The variance $Var(X) = E(X^2) - [E(X)]^2$.
$Var(X) = 3 - (\frac{3}{2})^2 = 3 - \frac{9}{4} = \frac{12 - 9}{4} = \frac{3}{4}$.

(A) The cumulative distribution function $F(x)$ is defined as $F(x) = \int_{-\infty}^{x} f(t) dt$.
For $0 < x < 4$,$F(x) = \int_{0}^{x} \frac{t}{8} dt$.
$F(x) = \frac{1}{8} \left[ \frac{t^2}{2} \right]_{0}^{x} = \frac{1}{16} (x^2 - 0) = \frac{x^2}{16}$.
Now,substitute $x = 0.5$ into the expression for $F(x)$:
$F(0.5) = \frac{(0.5)^2}{16} = \frac{0.25}{16} = \frac{1/4}{16} = \frac{1}{64}$.

(A) To find the variance of the random variable $X$,we use the formula $Var(X) = E(X^2) - [E(X)]^2$.

$x_{i}$	$p(x_{i})$	$p_{i} x_{i}$	$p_{i} x_{i}^{2}$
$0$	$\frac{1}{5}$	$0$	$0$
$1$	$\frac{2}{5}$	$\frac{2}{5}$	$\frac{2}{5}$
$2$	$\frac{2}{5}$	$\frac{4}{5}$	$\frac{8}{5}$

First,calculate the mean $E(X) = \sum p_{i} x_{i} = 0 + \frac{2}{5} + \frac{4}{5} = \frac{6}{5}$.
Next,calculate $E(X^2) = \sum p_{i} x_{i}^{2} = 0 + \frac{2}{5} + \frac{8}{5} = \frac{10}{5} = 2$.
Now,calculate the variance: $Var(X) = E(X^2) - [E(X)]^2 = 2 - (\frac{6}{5})^2 = 2 - \frac{36}{25} = \frac{50 - 36}{25} = \frac{14}{25}$.

(A) Given the p.d.f. of a random variable $X$: $f(x) = \begin{cases} 3(1 - 2x^2), & 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$
The probability is calculated as: $P\left(\frac{1}{4} < X < \frac{1}{3}\right) = \int_{1/4}^{1/3} 3(1 - 2x^2) \, dx$
$= 3 \left[ x - \frac{2}{3}x^3 \right]_{1/4}^{1/3}$
$= 3 \left[ \left( \frac{1}{3} - \frac{2}{3} \cdot \frac{1}{27} \right) - \left( \frac{1}{4} - \frac{2}{3} \cdot \frac{1}{64} \right) \right]$
$= 3 \left[ \left( \frac{1}{3} - \frac{2}{81} \right) - \left( \frac{1}{4} - \frac{1}{96} \right) \right]$
$= 3 \left[ \frac{27-2}{81} - \frac{24-1}{96} \right] = 3 \left[ \frac{25}{81} - \frac{23}{96} \right]$
$= 3 \left[ \frac{25 \times 32 - 23 \times 27}{2592} \right] = 3 \left[ \frac{800 - 621}{2592} \right]$
$= 3 \times \frac{179}{2592} = \frac{179}{864}$

(A) Given the p.d.f. of a random variable $x$ is $f(x) = \frac{1}{4a}$ for $0 < x < 4a$.
We are given the equation $P(x < \frac{3a}{2}) = k P(x > \frac{5a}{2})$.
Calculating the left side: $P(x < \frac{3a}{2}) = \int_{0}^{\frac{3a}{2}} \frac{1}{4a} dx = \frac{1}{4a} [x]_{0}^{\frac{3a}{2}} = \frac{1}{4a} \times \frac{3a}{2} = \frac{3}{8}$.
Calculating the right side: $P(x > \frac{5a}{2}) = \int_{\frac{5a}{2}}^{4a} \frac{1}{4a} dx = \frac{1}{4a} [x]_{\frac{5a}{2}}^{4a} = \frac{1}{4a} (4a - \frac{5a}{2}) = \frac{1}{4a} \times \frac{3a}{2} = \frac{3}{8}$.
Substituting these values into the given equation: $\frac{3}{8} = k \times \frac{3}{8}$.
Therefore,$k = 1$.

(A) The probability $P(a \leq x \leq b)$ is given by $F(b) - F(a)$.
Given $F(x) = \frac{x-25}{10}$.
We need to find $P(27 \leq x \leq 33) = F(33) - F(27)$.
$F(33) = \frac{33-25}{10} = \frac{8}{10} = 0.8$.
$F(27) = \frac{27-25}{10} = \frac{2}{10} = 0.2$.
Therefore,$P(27 \leq x \leq 33) = 0.8 - 0.2 = 0.6 = \frac{3}{5}$.

(A) For any probability distribution,the sum of all probabilities must be equal to $1$.
$\sum P(X=x) = K + 3K + 5K + 7K + 8K + K = 1$
$25K = 1 \implies K = \frac{1}{25}$
We need to find $P(2 \leq X < 5)$,which includes the values $X = 2, 3, 4$.
$P(2 \leq X < 5) = P(X=2) + P(X=3) + P(X=4)$
$= 3K + 5K + 7K = 15K$
Substituting $K = \frac{1}{25}$:
$= 15 \times \frac{1}{25} = \frac{15}{25} = \frac{3}{5}$

(D) When a coin is tossed three times,the total number of outcomes is $2^3 = 8$. The sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
Let $n(H)$ be the number of heads and $n(T)$ be the number of tails. $X = |n(H) - n(T)|$.
For $HHH: n(H)=3, n(T)=0, X=|3-0|=3$.
For $HHT: n(H)=2, n(T)=1, X=|2-1|=1$.
For $HTH: n(H)=2, n(T)=1, X=|2-1|=1$.
For $HTT: n(H)=1, n(T)=2, X=|1-2|=1$.
For $THH: n(H)=2, n(T)=1, X=|2-1|=1$.
For $THT: n(H)=1, n(T)=2, X=|1-2|=1$.
For $TTH: n(H)=1, n(T)=2, X=|1-2|=1$.
For $TTT: n(H)=0, n(T)=3, X=|0-3|=3$.
The outcomes where $X=1$ are $\{HHT, HTH, HTT, THH, THT, TTH\}$.
There are $6$ such outcomes.
Therefore,$P(X=1) = \frac{6}{8} = \frac{3}{4}$.

(B) The possible outcomes of rolling a die are $\{1, 2, 3, 4, 5, 6\}$. We determine the number of positive divisors for each outcome as follows:

Outcome	Positive Divisors	Number of Divisors $(X)$
$1$	$1$	$1$
$2$	$1, 2$	$2$
$3$	$1, 3$	$2$
$4$	$1, 2, 4$	$3$
$5$	$1, 5$	$2$
$6$	$1, 2, 3, 6$	$4$

The set of values taken by the random variable $X$ is the range. From the table,the values are $\{1, 2, 3, 4\}$. Thus,the range of $X$ is $\{1, 2, 3, 4\}$.

(A) The probability $P(3 < X \leq 5)$ is given by $F(5) - F(3)$.
From the given table:
$F(5) = 0.85$
$F(3) = 0.48$
Therefore,$P(3 < X \leq 5) = F(5) - F(3) = 0.85 - 0.48 = 0.37$.

(D) $x$ : number of defective pens.
Two pens are taken from the box.
Therefore,$x$ can take values $0, 1, 2$.
$P(x=0) = \frac{{}^4C_2}{{}^6C_2} = \frac{6}{15}$
$P(x=1) = \frac{{}^2C_1 \times {}^4C_1}{{}^6C_2} = \frac{8}{15}$
$P(x=2) = \frac{{}^2C_2}{{}^6C_2} = \frac{1}{15}$

$x_i$	$P(x_i)$	$x_i P(x_i)$	$x_i^2 P(x_i)$
$0$	$\frac{6}{15}$	$0$	$0$
$1$	$\frac{8}{15}$	$\frac{8}{15}$	$\frac{8}{15}$
$2$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{4}{15}$

$E(x) = \sum x_i P(x_i) = 0 + \frac{8}{15} + \frac{2}{15} = \frac{10}{15} = \frac{2}{3}$
$E(x^2) = \sum x_i^2 P(x_i) = 0 + \frac{8}{15} + \frac{4}{15} = \frac{12}{15} = \frac{4}{5}$
$\text{Variance}(x) = E(x^2) - [E(x)]^2 = \frac{4}{5} - (\frac{2}{3})^2 = \frac{4}{5} - \frac{4}{9} = \frac{36-20}{45} = \frac{16}{45}$
$\text{Standard Deviation} = \sqrt{\text{Variance}(x)} = \sqrt{\frac{16}{45}} = \frac{4}{3 \sqrt{5}}$

(C) The cumulative distribution function $F(x)$ is defined as $F(x) = P(X \leq x)$.
For $x = 0$,we have $F(0) = P(X \leq 0)$.
From the given probability distribution table:
$P(X \leq 0) = P(X = -2) + P(X = -1) + P(X = 0)$
$P(X \leq 0) = 0.2 + 0.5 + 0.15 = 0.85$.
Alternatively,we know that the sum of all probabilities is $1$,so $P(X \leq 0) + P(X > 0) = 1$.
Therefore,$F(0) = P(X \leq 0) = 1 - P(X > 0)$.
Thus,the correct option is $C$.

(B) The probability density function is given by $f(x) = \frac{1}{5}$ for $0 \leq x \leq 5$ and $0$ otherwise.
We need to find the probability that the waiting time $X$ is not more than $4$ minutes,which is $P(X \leq 4)$.
This is calculated by integrating the probability density function from the lower bound to $4$:
$P(X \leq 4) = \int_{0}^{4} f(x) \, dx$
$P(X \leq 4) = \int_{0}^{4} \frac{1}{5} \, dx$
$P(X \leq 4) = \left[ \frac{x}{5} \right]_{0}^{4}$
$P(X \leq 4) = \frac{4}{5} - \frac{0}{5} = \frac{4}{5} = 0.8$
Therefore,the probability is $0.8$.

(C) When two coins are tossed simultaneously,the sample space is $S = \{HH, HT, TH, TT\}$.
Let $X$ be the random variable representing the number of heads.
The possible values for $X$ are $0, 1, 2$.
The probabilities are:
$P(X=0) = P(TT) = \frac{1}{4}$
$P(X=1) = P(HT, TH) = \frac{2}{4} = \frac{1}{2}$
$P(X=2) = P(HH) = \frac{1}{4}$
The expected value $E(X)$ is given by $\sum x_i P(x_i) = 0 \times \frac{1}{4} + 1 \times \frac{1}{2} + 2 \times \frac{1}{4} = 0 + \frac{1}{2} + \frac{1}{2} = 1$.
Alternatively,since this follows a binomial distribution with $n=2$ and $p=\frac{1}{2}$,$E(X) = np = 2 \times \frac{1}{2} = 1$.

(B) Given,$E = \{x \text{ is a prime number}\} = \{2, 3, 5, 7\}$.
$P(E) = P(2) + P(3) + P(5) + P(7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$.
Given,$F = \{x < 4\} = \{1, 2, 3\}$.
$P(F) = P(1) + P(2) + P(3) = 0.15 + 0.23 + 0.12 = 0.50$.
The intersection $E \cap F = \{x \text{ is prime and } x < 4\} = \{2, 3\}$.
$P(E \cap F) = P(2) + P(3) = 0.23 + 0.12 = 0.35$.
Using the formula $P(E \cup F) = P(E) + P(F) - P(E \cap F)$:
$P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$.

(C) The expected value $E(X)$ is calculated as follows:
$E(X) = \sum x_i P(x_i)$
$E(X) = (-2)(0.1) + (-1)(0.2) + (0)(0.2) + (1)(0.3) + (2)(0.15) + (3)(0.05)$
$E(X) = -0.2 - 0.2 + 0 + 0.3 + 0.3 + 0.15 = 0.35$
The expected value of the square $E(X^2)$ is:
$E(X^2) = \sum x_i^2 P(x_i)$
$E(X^2) = (-2)^2(0.1) + (-1)^2(0.2) + (0)^2(0.2) + (1)^2(0.3) + (2)^2(0.15) + (3)^2(0.05)$
$E(X^2) = (4)(0.1) + (1)(0.2) + 0 + (1)(0.3) + (4)(0.15) + (9)(0.05)$
$E(X^2) = 0.4 + 0.2 + 0 + 0.3 + 0.6 + 0.45 = 1.95$
The variance $Var(X)$ is given by:
$Var(X) = E(X^2) - [E(X)]^2$
$Var(X) = 1.95 - (0.35)^2$
$Var(X) = 1.95 - 0.1225 = 1.8275$

(C) The random variable $X$ takes values $x_i \in \{0, 1, 2\}$.
Given the mean $E(X) = \sum x_i P(x_i) = 1.2$.
Substituting the values: $(0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2)) = 1.2$.
Since $P(X=0) = 0.3$,we have: $0 + P(X=1) + 2P(X=2) = 1.2 \implies P(X=1) + 2P(X=2) = 1.2$ (Equation $1$).
Also,the sum of probabilities is $1$: $P(X=0) + P(X=1) + P(X=2) = 1$.
$0.3 + P(X=1) + P(X=2) = 1 \implies P(X=1) + P(X=2) = 0.7$ (Equation $2$).
Subtracting Equation $2$ from Equation $1$: $(P(X=1) + 2P(X=2)) - (P(X=1) + P(X=2)) = 1.2 - 0.7$.
$P(X=2) = 0.5$.
Substituting $P(X=2) = 0.5$ into Equation $2$: $P(X=1) + 0.5 = 0.7 \implies P(X=1) = 0.2$.

(B) The probability distribution of $X$ is given by:
| $x_i$ | $p_i$ | $x_i p_i$ | $x_i^2 p_i$ |
|---|---|---|---|
| $0$ | $0.4$ | $0$ | $0$ |
| $1$ | $0.6$ | $0.6$ | $0.6$ |
| Total | | $0.6$ | $0.6$ |
We know that $\text{Var}(X) = E(X^2) - [E(X)]^2$.
From the table,$E(X) = \sum x_i p_i = 0.6$.
$E(X^2) = \sum x_i^2 p_i = 0.6$.
Therefore,$\text{Var}(X) = 0.6 - (0.6)^2 = 0.6 - 0.36 = 0.24$.

(A) The expected daily demand (mean) is given by $E(X) = \Sigma P_i x_i$.
$E(X) = (5 \times 0.07) + (6 \times 0.2) + (7 \times 0.3) + (8 \times 0.3) + (9 \times 0.07) + (10 \times 0.06)$
$E(X) = 0.35 + 1.2 + 2.1 + 2.4 + 0.63 + 0.6 = 7.28$.
The variance is given by $Var(X) = \Sigma P_i x_i^2 - (E(X))^2$.
First,calculate $\Sigma P_i x_i^2$:
$\Sigma P_i x_i^2 = (25 \times 0.07) + (36 \times 0.2) + (49 \times 0.3) + (64 \times 0.3) + (81 \times 0.07) + (100 \times 0.06)$
$\Sigma P_i x_i^2 = 1.75 + 7.2 + 14.7 + 19.2 + 5.67 + 6.0 = 54.52$.
Now,calculate variance:
$Var(X) = 54.52 - (7.28)^2$
$Var(X) = 54.52 - 52.9984 \approx 1.52$.
Thus,the expected demand is $7.28$ and the variance is $1.52$.

(D) The expected value $E(X^2)$ is calculated using the formula $E(X^2) = \sum x_i^2 P(x_i)$.
Given the values:
$x_1 = 1, P(x_1) = \frac{1}{10} \implies x_1^2 P(x_1) = 1^2 \times \frac{1}{10} = \frac{1}{10}$
$x_2 = 2, P(x_2) = \frac{1}{5} \implies x_2^2 P(x_2) = 2^2 \times \frac{1}{5} = 4 \times \frac{1}{5} = \frac{4}{5} = \frac{8}{10}$
$x_3 = 3, P(x_3) = \frac{3}{10} \implies x_3^2 P(x_3) = 3^2 \times \frac{3}{10} = 9 \times \frac{3}{10} = \frac{27}{10}$
$x_4 = 4, P(x_4) = \frac{2}{5} \implies x_4^2 P(x_4) = 4^2 \times \frac{2}{5} = 16 \times \frac{2}{5} = \frac{32}{5} = \frac{64}{10}$
Summing these values:
$E(X^2) = \frac{1}{10} + \frac{8}{10} + \frac{27}{10} + \frac{64}{10} = \frac{1 + 8 + 27 + 64}{10} = \frac{100}{10} = 10$.
Therefore,the correct option is $D$.

(D) For a probability distribution,the sum of all probabilities must be equal to $1$.
Thus,$\sum_{x=1}^{4} P(x) = 1$.
Given $P(x) = \frac{c}{3} \binom{4}{x}$,we have:
$\frac{c}{3} [\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4}] = 1$.
We know that $\sum_{x=0}^{n} \binom{n}{x} = 2^n$.
Therefore,$\binom{4}{0} + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 2^4 = 16$.
Since $\binom{4}{0} = 1$,we have $\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 16 - 1 = 15$.
Substituting this into the equation:
$\frac{c}{3} \times 15 = 1$.
$5c = 1$.
$c = \frac{1}{5}$.

(D) For a probability distribution,the sum of all probabilities must be equal to $1$.
Thus,$\sum_{x=0}^{4} P(x) = 1$.
Substituting the given expression: $\sum_{x=0}^{4} C \binom{4}{x} = 1$.
$C \sum_{x=0}^{4} \binom{4}{x} = 1$.
We know that the sum of binomial coefficients $\sum_{k=0}^{n} \binom{n}{k} = 2^n$.
For $n = 4$,$\sum_{x=0}^{4} \binom{4}{x} = 2^4 = 16$.
Therefore,$C \times 16 = 1$.
$C = \frac{1}{16}$.

(A) We know that for a probability distribution,the sum of all probabilities must be equal to $1$,i.e.,$\sum P(X) = 1$.
Substituting the given values:
$0.3 + k + 2k + 2k = 1$
$0.3 + 5k = 1$
$5k = 1 - 0.3$
$5k = 0.7$
$k = \frac{0.7}{5}$
$k = 0.14$

(A) We know that for any probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$\sum P(x) = 1$.
Substituting the given values from the table:
$0.2 + k + k + 2k = 1$
Combining the like terms:
$0.2 + 4k = 1$
Subtracting $0.2$ from both sides:
$4k = 1 - 0.2$
$4k = 0.8$
Dividing by $4$:
$k = \frac{0.8}{4} = 0.2$
Thus,the value of $k$ is $0.2$.

(B) We know that the sum of probabilities in a probability distribution is $1$. Therefore,$\frac{25}{36} + k + \frac{1}{36} = 1 \Rightarrow k + \frac{26}{36} = 1 \Rightarrow k = 1 - \frac{13}{18} = \frac{5}{18}$.
The mean of a random variable $X$ is given by $E(X) = \sum x_i P(x_i)$.
Given $E(X) = \frac{1}{3}$,we have:
$E(X) = 0 \times \frac{25}{36} + 1 \times k + 2 \times \frac{1}{36} = k + \frac{2}{36} = k + \frac{1}{18}$.
Since $E(X) = \frac{1}{3}$,we have $k + \frac{1}{18} = \frac{1}{3} \Rightarrow k = \frac{1}{3} - \frac{1}{18} = \frac{6-1}{18} = \frac{5}{18}$.
The variance of $X$ is given by $Var(X) = E(X^2) - [E(X)]^2$.
First,calculate $E(X^2) = \sum x_i^2 P(x_i)$:
$E(X^2) = 0^2 \times \frac{25}{36} + 1^2 \times k + 2^2 \times \frac{1}{36} = 0 + k + \frac{4}{36} = k + \frac{1}{9}$.
Substituting $k = \frac{5}{18}$:
$E(X^2) = \frac{5}{18} + \frac{1}{9} = \frac{5+2}{18} = \frac{7}{18}$.
Now,$Var(X) = \frac{7}{18} - (\frac{1}{3})^2 = \frac{7}{18} - \frac{1}{9} = \frac{7-2}{18} = \frac{5}{18}$.