Probability distribution Questions in English

(B) The number of errors per page follows a Poisson distribution with parameter $\lambda = \frac{60}{600} = 0.1$.
The probability mass function is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!} = \frac{e^{-0.1} (0.1)^x}{x!}$.
We need to find the probability that a page contains at most two errors,which is $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=0) = \frac{e^{-0.1} (0.1)^0}{0!} = e^{-0.1}$.
$P(X=1) = \frac{e^{-0.1} (0.1)^1}{1!} = 0.1 e^{-0.1}$.
$P(X=2) = \frac{e^{-0.1} (0.1)^2}{2!} = \frac{0.01}{2} e^{-0.1} = 0.005 e^{-0.1}$.
Adding these probabilities: $P(X \le 2) = e^{-0.1} (1 + 0.1 + 0.005) = e^{-0.1} (1.105) = e^{-0.1} \left(\frac{1105}{1000}\right) = e^{-0.1} \left(\frac{221}{200}\right) = \frac{1}{e^{0.1}} \left(\frac{221}{200}\right)$.

(B) Given,the average number of births in a week is $35$.
Since there are $7$ days in a week,the average number of births in a day is $\lambda = \frac{35}{7} = 5$.
We use the Poisson distribution formula $P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$.
We need to find the probability of less than $3$ births in a day,which is $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$.
$P(X = 0) = \frac{5^0 e^{-5}}{0!} = e^{-5}$.
$P(X = 1) = \frac{5^1 e^{-5}}{1!} = 5e^{-5}$.
$P(X = 2) = \frac{5^2 e^{-5}}{2!} = \frac{25}{2}e^{-5}$.
Adding these probabilities: $P(X < 3) = e^{-5} + 5e^{-5} + \frac{25}{2}e^{-5} = e^{-5}(1 + 5 + 12.5) = 18.5e^{-5} = \frac{37}{2e^5}$.

(B) Given the probability function $P(X=r) = r/k$ for $r = 1, 2, 3, 4, 5$.
Since the sum of all probabilities must be $1$,we have:
$\sum_{r=1}^{5} P(X=r) = 1 \Rightarrow \frac{1}{k} + \frac{2}{k} + \frac{3}{k} + \frac{4}{k} + \frac{5}{k} = 1$
$\frac{1+2+3+4+5}{k} = 1 \Rightarrow \frac{15}{k} = 1 \Rightarrow k = 15$.
We need to find $P(X=2 \text{ or } X=k/3)$.
Since $k=15$,$k/3 = 15/3 = 5$.
So,$P(X=2 \text{ or } X=5) = P(X=2) + P(X=5) = \frac{2}{15} + \frac{5}{15} = \frac{7}{15}$.
Now,let us check the options:
Option $B$: $P(X=4 \text{ or } X=k/5) = P(X=4 \text{ or } X=15/5) = P(X=4 \text{ or } X=3) = \frac{4}{15} + \frac{3}{15} = \frac{7}{15}$.
Thus,$P(X=2 \text{ or } X=k/3) = P(X=4 \text{ or } X=k/5)$.

(D) When two dice are rolled,the total number of outcomes is $6 \times 6 = 36$. Let $X$ be the absolute difference of the numbers on the two dice. The possible values of $X$ are $0, 1, 2, 3, 4, 5$. The probability distribution is as follows:
| $X$ | $P(X)$ | $P_i X_i$ |
|---|---|---|
| $0$ | $6/36$ | $0$ |
| $1$ | $10/36$ | $10/36$ |
| $2$ | $8/36$ | $16/36$ |
| $3$ | $6/36$ | $18/36$ |
| $4$ | $4/36$ | $16/36$ |
| $5$ | $2/36$ | $10/36$ |
The mean $\mu$ is given by $\sum P_i X_i$:
$\mu = 0 + \frac{10}{36} + \frac{16}{36} + \frac{18}{36} + \frac{16}{36} + \frac{10}{36}$
$\mu = \frac{10 + 16 + 18 + 16 + 10}{36} = \frac{70}{36} = \frac{35}{18}$.

(C) Given,$n = 50$. Let $p$ be the probability of success in one trial. Then the parameter of the Poisson distribution is $\lambda = np = 50p$.
The probability mass function for a Poisson variable is $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
Given the equation: $2 P(X=1) = 5 P(X=5) + 2 P(X=3)$.
Substituting the formula: $2 \frac{e^{-\lambda} \lambda^1}{1!} = 5 \frac{e^{-\lambda} \lambda^5}{5!} + 2 \frac{e^{-\lambda} \lambda^3}{3!}$.
Dividing both sides by $e^{-\lambda}$ (since $e^{-\lambda} \neq 0$): $2 \lambda = 5 \frac{\lambda^5}{120} + 2 \frac{\lambda^3}{6}$.
$2 \lambda = \frac{\lambda^5}{24} + \frac{\lambda^3}{3}$.
Multiplying by $24$: $48 \lambda = \lambda^5 + 8 \lambda^3$.
Since $\lambda \neq 0$,divide by $\lambda$: $\lambda^4 + 8 \lambda^2 - 48 = 0$.
Let $u = \lambda^2$,then $u^2 + 8u - 48 = 0$.
$(u + 12)(u - 4) = 0$.
So,$\lambda^2 = 4$ or $\lambda^2 = -12$. Since $\lambda^2$ must be positive,$\lambda^2 = 4$,which gives $\lambda = 2$.
Finally,$p = \frac{\lambda}{n} = \frac{2}{50} = 0.04$.

(C) First,we calculate $E(X)$ and $E(X^2)$ using the given distribution:
$E(X) = \Sigma x P(X=x) = (-1)(\frac{1}{3}) + (0)(\frac{1}{6}) + (1)(\frac{1}{6}) + (2)(\frac{1}{3}) = -\frac{1}{3} + 0 + \frac{1}{6} + \frac{2}{3} = \frac{-2+0+1+4}{6} = \frac{3}{6} = \frac{1}{2}$
$E(X^2) = \Sigma x^2 P(X=x) = (-1)^2(\frac{1}{3}) + (0)^2(\frac{1}{6}) + (1)^2(\frac{1}{6}) + (2)^2(\frac{1}{3}) = \frac{1}{3} + 0 + \frac{1}{6} + \frac{4}{3} = \frac{2+0+1+8}{6} = \frac{11}{6}$
Now,we find $\operatorname{Var}(X) = E(X^2) - (E(X))^2 = \frac{11}{6} - (\frac{1}{2})^2 = \frac{11}{6} - \frac{1}{4} = \frac{22-3}{12} = \frac{19}{12}$
Finally,we calculate $6 E(X^2) - \operatorname{Var}(X)$:
$6 E(X^2) - \operatorname{Var}(X) = 6(\frac{11}{6}) - \frac{19}{12} = 11 - \frac{19}{12} = \frac{132-19}{12} = \frac{113}{12}$

(A) Let $N$ be the number appearing on the die. The possible values for $N$ are $\{1, 2, 3, 4, 5, 6\}$.
If $N$ is even $(N \in \{2, 4, 6\})$,the number of chocolates $X = N + 2$.
For $N = 2$,$X = 2 + 2 = 4$.
For $N = 4$,$X = 4 + 2 = 6$.
For $N = 6$,$X = 6 + 2 = 8$.
If $N$ is odd $(N \in \{1, 3, 5\})$,the number of chocolates $X = N + 3$.
For $N = 1$,$X = 1 + 3 = 4$.
For $N = 3$,$X = 3 + 3 = 6$.
For $N = 5$,$X = 5 + 3 = 8$.
Thus,the set of all possible values of $X$ is $\{4, 6, 8\}$.
Therefore,the range of $X$ is $\{4, 6, 8\}$.

(A) When two dice are rolled,the sum $X$ can take values from $2$ to $12$. The probability distribution is as follows:
$X: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$
$P(X): \frac{1}{36}, \frac{2}{36}, \frac{3}{36}, \frac{4}{36}, \frac{5}{36}, \frac{6}{36}, \frac{5}{36}, \frac{4}{36}, \frac{3}{36}, \frac{2}{36}, \frac{1}{36}$
The mean $\mu = E(X) = \sum X_i P(X_i) = \frac{2(1)+3(2)+4(3)+5(4)+6(5)+7(6)+8(5)+9(4)+10(3)+11(2)+12(1)}{36} = \frac{252}{36} = 7$.
Now,calculate the probabilities:
$P(X < 5) = P(X=2) + P(X=3) + P(X=4) = \frac{1+2+3}{36} = \frac{6}{36}$.
$P(X > 9) = P(X=10) + P(X=11) + P(X=12) = \frac{3+2+1}{36} = \frac{6}{36}$.
$P(X = 7) = \frac{6}{36}$.
Substituting these values into the expression $\mu + P(X < 5) + P(X > 9) + P(X = 7)$:
$= 7 + \frac{6}{36} + \frac{6}{36} + \frac{6}{36} = 7 + \frac{18}{36} = 7 + \frac{1}{2} = \frac{15}{2}$.

(B) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$.
Given $P(X=2) = P(X=3)$,we have:
$\frac{\lambda^2 e^{-\lambda}}{2!} = \frac{\lambda^3 e^{-\lambda}}{3!}$
$\frac{\lambda^2}{2} = \frac{\lambda^3}{6}$
Since $\lambda > 0$,we can divide by $\lambda^2$:
$\frac{1}{2} = \frac{\lambda}{6} \Rightarrow \lambda = 3$.
Now,we calculate $P(X=5)$:
$P(X=5) = \frac{3^5 e^{-3}}{5!} = \frac{243 e^{-3}}{120} = \frac{81 e^{-3}}{40} = \frac{81}{40 e^3}$.
Thus,option $B$ is correct.

(A) Given $2 P(X=1) = 3 P(X=2) = P(X=3) = 5 P(X=4) = k$.
Then $P(X=1) = \frac{k}{2}, P(X=2) = \frac{k}{3}, P(X=3) = k, P(X=4) = \frac{k}{5}$.
Since $\sum P(X) = 1$,we have $\frac{k}{2} + \frac{k}{3} + k + \frac{k}{5} = 1$.
$\Rightarrow k(\frac{15+10+30+6}{30}) = 1 \Rightarrow k(\frac{61}{30}) = 1 \Rightarrow k = \frac{30}{61}$.
The probability distribution is:

$x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{15}{61}$	$\frac{10}{61}$	$\frac{30}{61}$	$\frac{6}{61}$

Mean $\mu = E(X) = \sum x P(x) = 1(\frac{15}{61}) + 2(\frac{10}{61}) + 3(\frac{30}{61}) + 4(\frac{6}{61}) = \frac{15+20+90+24}{61} = \frac{149}{61}$.
$E(X^2) = \sum x^2 P(x) = 1^2(\frac{15}{61}) + 2^2(\frac{10}{61}) + 3^2(\frac{30}{61}) + 4^2(\frac{6}{61}) = \frac{15+40+270+96}{61} = \frac{421}{61}$.
Variance $\sigma^2 = E(X^2) - \mu^2$.
We need $\sigma^2 + \mu^2 = E(X^2) - \mu^2 + \mu^2 = E(X^2)$.
Therefore,$\sigma^2 + \mu^2 = \frac{421}{61}$.

(A) For a Poisson distribution,the probability mass function is given by $P(X=r) = \frac{\lambda^r e^{-\lambda}}{r!}$,where $\lambda$ is the parameter of the distribution.
Given that $P(X=1) = 3P(X=2)$.
Substituting the formula:
$\frac{\lambda^1 e^{-\lambda}}{1!} = 3 \times \frac{\lambda^2 e^{-\lambda}}{2!}$
$\lambda = 3 \times \frac{\lambda^2}{2}$
Since $\lambda \neq 0$,we divide by $\lambda$:
$1 = \frac{3\lambda}{2} \implies \lambda = \frac{2}{3}$.
Now,we calculate $P(X=3)$:
$P(X=3) = \frac{\lambda^3 e^{-\lambda}}{3!} = \frac{(\frac{2}{3})^3 e^{-\frac{2}{3}}}{6}$
$P(X=3) = \frac{\frac{8}{27} e^{-\frac{2}{3}}}{6} = \frac{8}{27 \times 6} e^{-\frac{2}{3}} = \frac{4}{81} e^{-\frac{2}{3}}$.

(C) For a probability distribution,the sum of all probabilities must be $1$:
$\sum P(X=x) = a + a + a + b + b + 0.3 = 1$
$3a + 2b + 0.3 = 1$
$3a + 2b = 0.7$ --- $(i)$
The mean of a random variable $X$ is given by $E(X) = \sum x_i P(x_i) = 4.2$:
$1(a) + 2(a) + 3(a) + 4(b) + 5(b) + 6(0.3) = 4.2$
$a + 2a + 3a + 4b + 5b + 1.8 = 4.2$
$6a + 9b = 4.2 - 1.8$
$6a + 9b = 2.4$
Dividing by $3$,we get:
$2a + 3b = 0.8$ --- $(ii)$
Now,solve the system of linear equations $(i)$ and $(ii)$:
Multiply $(i)$ by $3$ and $(ii)$ by $2$:
$9a + 6b = 2.1$ --- $(iii)$
$4a + 6b = 1.6$ --- $(iv)$
Subtract $(iv)$ from $(iii)$:
$5a = 0.5 \Rightarrow a = 0.1$
Substitute $a = 0.1$ into $(i)$:
$3(0.1) + 2b = 0.7$
$0.3 + 2b = 0.7$
$2b = 0.4 \Rightarrow b = 0.2$
Thus,$a = 0.1$ and $b = 0.2$.

(C) To find the variance of the random variable $X$,we use the formula: $\text{Var}(X) = E(X^2) - [E(X)]^2$.
First,we calculate the mean $E(X) = \sum P_i X_i$:
$E(X) = (0 \times 0.1) + (1 \times 0.4) + (2 \times 0.3) + (3 \times 0.2) + (4 \times 0) = 0 + 0.4 + 0.6 + 0.6 + 0 = 1.6$.
Next,we calculate $E(X^2) = \sum P_i X_i^2$:
$E(X^2) = (0^2 \times 0.1) + (1^2 \times 0.4) + (2^2 \times 0.3) + (3^2 \times 0.2) + (4^2 \times 0) = 0 + 0.4 + 1.2 + 1.8 + 0 = 3.4$.
Now,calculate the variance:
$\text{Var}(X) = 3.4 - (1.6)^2 = 3.4 - 2.56 = 0.84$.

(B) For a probability distribution,the sum of all probabilities must be $1$:
$\Sigma P(X=x) = K + 2K + 3K + 2K + K = 9K = 1$
$\therefore K = \frac{1}{9}$
The mean $E(X) = \Sigma x_i P(x_i) = (1 \times K) + (2 \times 2K) + (3 \times 3K) + (4 \times 2K) + (5 \times K)$
$= K + 4K + 9K + 8K + 5K = 27K = 27 \times \frac{1}{9} = 3$
$E(X^2) = \Sigma x_i^2 P(x_i) = (1^2 \times K) + (2^2 \times 2K) + (3^2 \times 3K) + (4^2 \times 2K) + (5^2 \times K)$
$= K + 8K + 27K + 32K + 25K = 93K = 93 \times \frac{1}{9} = \frac{93}{9} = \frac{31}{3}$
Variance $Var(X) = E(X^2) - [E(X)]^2$
$= \frac{31}{3} - (3)^2 = \frac{31}{3} - 9 = \frac{31 - 27}{3} = \frac{4}{3}$

(B) Given that $X$ is a Poisson variate with parameter $\lambda$,the probability mass function is $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
Given $\alpha = P(X=1) = P(X=2)$:
$\frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!}$
$\lambda = \frac{\lambda^2}{2} \Rightarrow \lambda = 2$ (since $\lambda > 0$).
Now,calculate $\alpha$:
$\alpha = P(X=1) = \frac{e^{-2} \times 2^1}{1!} = 2e^{-2}$.
We need to find $P(X=4)$:
$P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!} = \frac{e^{-2} \times 2^4}{24} = \frac{16 e^{-2}}{24} = \frac{2}{3} e^{-2}$.
Since $\alpha = 2e^{-2}$,we have $e^{-2} = \frac{\alpha}{2}$.
Substituting this into the expression for $P(X=4)$:
$P(X=4) = \frac{2}{3} \times \frac{\alpha}{2} = \frac{\alpha}{3}$.

(C) Let $\lambda$ be the mean of the Poisson distribution for the random variable $X$.
The probability mass function is given by $P(X=r) = \frac{\lambda^r e^{-\lambda}}{r!}$ for $r = 0, 1, 2, \dots$.
Given $P(X=1) = P(X=2)$,we have:
$\frac{\lambda^1 e^{-\lambda}}{1!} = \frac{\lambda^2 e^{-\lambda}}{2!}$
$\lambda = \frac{\lambda^2}{2}$
Since $\lambda > 0$,we divide by $\lambda$ to get $1 = \frac{\lambda}{2}$,which implies $\lambda = 2$.
Now,we calculate $P(X=5)$:
$P(X=5) = \frac{2^5 e^{-2}}{5!} = \frac{32 e^{-2}}{120}$.
Simplifying the fraction $\frac{32}{120}$ by dividing both numerator and denominator by $8$,we get $\frac{4}{15}$.
Thus,$P(X=5) = \frac{4}{15} e^{-2}$.

(C) For a Poisson distribution,the probability mass function is given by $P(X=x) = \frac{e^{-m} m^x}{x!}$,where $m$ is the parameter (mean and variance).
Given $P(X=0) = 0.8$.
Substituting $x=0$ in the formula:
$P(X=0) = \frac{e^{-m} m^0}{0!} = e^{-m} = 0.8$.
Taking the natural logarithm on both sides:
$-m = \ln(0.8) = \ln(\frac{8}{10}) = \ln(\frac{4}{5})$.
Therefore,$m = -\ln(\frac{4}{5}) = \ln((\frac{4}{5})^{-1}) = \ln(\frac{5}{4}) = \ln(1.25)$.
Since the variance of a Poisson distribution is equal to its parameter $m$,the variance is $\ln(1.25)$ or $\log _e 1.25$.

(D) The sum of probabilities in a probability distribution is always $1$.
Given $P(X=0) = 0.4$.
Since $P(X=0) + P(X=1) + P(X=2) = 1$,we have $0.4 + P(X=1) + P(X=2) = 1$.
$P(X=1) + P(X=2) = 0.6$.
Given $P(X=1) = P(X=2)$,let $P(X=1) = P(X=2) = p$.
Then $p + p = 0.6 \Rightarrow 2p = 0.6 \Rightarrow p = 0.3$.
So,$P(X=1) = 0.3$ and $P(X=2) = 0.3$.
The mean $E(X)$ is given by $\sum x_i P(x_i)$.
$E(X) = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2))$.
$E(X) = (0 \times 0.4) + (1 \times 0.3) + (2 \times 0.3)$.
$E(X) = 0 + 0.3 + 0.6 = 0.9$.

(C) Given that the mean of the Poisson distribution is $\lambda = \frac{1}{2}$.
The probability mass function of a Poisson distribution is given by $P(X=n) = \frac{\lambda^n e^{-\lambda}}{n!}$.
We need to find the ratio $\frac{P(X=3)}{P(X=2)}$.
$P(X=3) = \frac{(\frac{1}{2})^3 e^{-1/2}}{3!}$ and $P(X=2) = \frac{(\frac{1}{2})^2 e^{-1/2}}{2!}$.
Taking the ratio:
$\frac{P(X=3)}{P(X=2)} = \frac{\frac{(\frac{1}{2})^3 e^{-1/2}}{3!}}{\frac{(\frac{1}{2})^2 e^{-1/2}}{2!}}$
$= \frac{(\frac{1}{2})^3}{3!} \times \frac{2!}{(\frac{1}{2})^2}$
$= \frac{1}{2} \times \frac{2!}{3!}$
$= \frac{1}{2} \times \frac{2}{6} = \frac{1}{6}$.
Thus,the ratio is $1:6$.

(C) The total number of outcomes when a die is rolled $12$ times is $6^{12}$.
We want each of the $6$ faces to appear exactly $2$ times.
This is a multinomial distribution problem where we arrange $12$ items into $6$ groups of size $2$ each.
The number of ways to distribute $12$ outcomes such that each face appears twice is given by the multinomial coefficient:
$\frac{12!}{2! 2! 2! 2! 2! 2!} = \frac{12!}{(2!)^6} = \frac{12!}{2^6}$.
Therefore,the required probability is:
$P = \frac{12!}{2^6 \times 6^{12}}$.

(B) Let $q = 1-p$ be the probability of getting a tail. The event that the first head appears on an even toss means the sequence of outcomes is $(T, H), (T, T, T, H), (T, T, T, T, T, H), \dots$
The probability of this occurring is $P = qp + q^3p + q^5p + \dots$
This is an infinite geometric series with first term $a = qp$ and common ratio $r = q^2$.
Since $0 < p < 1$,we have $0 < q < 1$,so $|q^2| < 1$.
The sum of the series is $P = \frac{a}{1-r} = \frac{qp}{1-q^2}$.
Given $P = \frac{2}{5}$,we have $\frac{qp}{1-q^2} = \frac{2}{5}$.
Substituting $q = 1-p$,we get $\frac{(1-p)p}{1-(1-p)^2} = \frac{2}{5}$.
Simplifying the denominator: $1 - (1 - 2p + p^2) = 2p - p^2 = p(2-p)$.
So,$\frac{p(1-p)}{p(2-p)} = \frac{2}{5}$.
Canceling $p$ (since $p \neq 0$),we get $\frac{1-p}{2-p} = \frac{2}{5}$.
Cross-multiplying: $5(1-p) = 2(2-p) \Rightarrow 5 - 5p = 4 - 2p$.
$1 = 3p \Rightarrow p = \frac{1}{3}$.

(C) Given the expected value $E(X) = \sum X_i P(X_i) = \frac{263}{15}$.
Calculating the sum: $E(X) = (4k \cdot \frac{2}{15}) + (\frac{30}{7}k \cdot \frac{1}{15}) + (\frac{32}{7}k \cdot \frac{2}{15}) + (\frac{34}{7}k \cdot \frac{1}{5}) + (\frac{36}{7}k \cdot \frac{1}{15}) + (\frac{38}{7}k \cdot \frac{2}{15}) + (\frac{40}{7}k \cdot \frac{1}{5}) + (6k \cdot \frac{1}{15})$.
$E(X) = \frac{k}{105} [ (28 \cdot 2) + (30 \cdot 1) + (32 \cdot 2) + (34 \cdot 3) + (36 \cdot 1) + (38 \cdot 2) + (40 \cdot 3) + (42 \cdot 1) ] = \frac{k}{105} [ 56 + 30 + 64 + 102 + 36 + 76 + 120 + 42 ] = \frac{526k}{105} = \frac{263}{15}$.
Solving for $k$: $k = \frac{263}{15} \cdot \frac{105}{526} = \frac{263}{526} \cdot \frac{105}{15} = \frac{1}{2} \cdot 7 = \frac{7}{2}$.
Substituting $k = \frac{7}{2}$ into the values of $X$: $X$ takes values ${14, 15, 16, 17, 18, 19, 20, 21}$.
We need $P(X < 20) = P(X=14) + P(X=15) + P(X=16) + P(X=17) + P(X=18) + P(X=19)$.
$P(X < 20) = \frac{2}{15} + \frac{1}{15} + \frac{2}{15} + \frac{1}{5} + \frac{1}{15} + \frac{2}{15} = \frac{2+1+2+3+1+2}{15} = \frac{11}{15}$.

(C) The sum of probabilities must be $1$:
$\frac{2a + 1}{30} + \frac{8a - 1}{30} + \frac{4a + 1}{30} + b = 1$
$\frac{14a + 1}{30} + b = 1 \Rightarrow 14a + 30b = 29 \dots (1)$
We are given $\sigma^{2} + \mu^{2} = 2$. Since $\sigma^{2} = E[X^{2}] - \mu^{2}$,we have $E[X^{2}] = 2$.
$E[X^{2}] = \sum x_{i}^{2} p(x_{i}) = 0^{2} \cdot \frac{2a+1}{30} + 1^{2} \cdot \frac{8a-1}{30} + 2^{2} \cdot \frac{4a+1}{30} + 3^{2} \cdot b = 2$
$\frac{8a - 1 + 16a + 4}{30} + 9b = 2$
$\frac{24a + 3}{30} + 9b = 2$ $\Rightarrow 24a + 3 + 270b = 60$ $\Rightarrow 24a + 270b = 57$
Dividing by $3$: $8a + 90b = 19 \dots (2)$
From $(1)$,$30b = 29 - 14a$. Substitute into $(2)$:
$8a + 3(29 - 14a) = 19$
$8a + 87 - 42a = 19$
$-34a = -68 \Rightarrow a = 2$
Substituting $a = 2$ into $(1)$: $14(2) + 30b = 29$ $\Rightarrow 28 + 30b = 29$ $\Rightarrow 30b = 1$ $\Rightarrow b = \frac{1}{30}$
Therefore,$\frac{a}{b} = \frac{2}{1/30} = 60$.

(D) For a probability distribution,the sum of all probabilities must be equal to $1$.
$\sum P(x_i) = 0 + 2k + k + 3k + 2k^2 + 2k + (k^2 + k) + 7k^2 = 1$
Combining the terms,we get: $10k^2 + 9k = 1$
Rearranging into a quadratic equation: $10k^2 + 9k - 1 = 0$
Factoring the equation: $(10k - 1)(k + 1) = 0$
Since $k$ must be positive for probabilities to be non-negative,we have $k = \frac{1}{10} = 0.1$.
We need to find $P(3 < x \leq 6) = P(x=4) + P(x=5) + P(x=6)$.
$P(3 < x \leq 6) = 2k^2 + 2k + (k^2 + k) = 3k^2 + 3k$.
Substituting $k = 0.1$:
$P(3 < x \leq 6) = 3(0.1)^2 + 3(0.1) = 3(0.01) + 0.3 = 0.03 + 0.3 = 0.33$.