Got it — I’ll work through every part and give clean, exam-ready answers.
(Where a test is asked for, I show the test statistic, d.f., the critical value, and the decision.)

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1. (Attempt any two)

1(a) Two unbiased dice → expected sum

Let $X_1,X_2\sim \text{Unif}\{1,2,\ldots,6\}$ i.i.d.
$\mathbb E[X_i]=\frac{1+2+\cdots+6}{6}=3.5$.
Sum $S=X_1+X_2\Rightarrow \mathbb E[S]=\mathbb E[X_1]+\mathbb E[X_2]=7$.
(Variance if required: $\operatorname{Var}(S)=2\operatorname{Var}(X_1)=2\cdot \frac{35}{12}=\frac{35}{6}$.)

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1(b) Fit a binomial distribution to the 4-coin data

Hypothesis: coins unbiased ⇒ $n=4,\;p=\tfrac12$.
$P(R=r)=\binom{4}{r}2^{-4}=\binom{4}{r}/16$.
Expected frequency in 160 trials: $E_r=160\,P(R=r)=10\binom{4}{r}$.

$r$	0	1	2	3	4
Observed $O_r$	8	34	69	43	6
Expected $E_r$	10	40	60	40	10

(Optionally, χ² GOF test:
$\chi^2=\sum (O_r-E_r)^2/E_r=4.475$ with $k-1=4$ d.f.; critical at 5% $=9.488$.
Since $4.475<9.488$, the binomial $\text{Bin}(4,\tfrac12)$ fits.)

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1(c) Poisson process and the Poisson distribution

Poisson process $\{N(t),t\ge0\}$ with rate $\lambda>0$:
(i) $N(0)=0$.
(ii) Independent increments.
(iii) Stationary increments: $N(t+h)-N(t)\sim\text{Poisson}(\lambda h)$.
(iv) $\Pr(N(h)=1)=\lambda h+o(h)$, $\Pr(N(h)\ge2)=o(h)$ as $h\downarrow0$.

Derivation. Partition $[0,t]$ into $n$ pieces of length $h=t/n$ and use (ii)–(iv). The p.g.f./mgf limit or the binomial limit gives

$$\Pr\{N(t)=k\}=\frac{e^{-\lambda t}(\lambda t)^k}{k!},\quad k=0,1,2,\ldots$$

This is the Poisson distribution with mean and variance both $\lambda t$.
Example: number of calls arriving at a switchboard per minute.

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2. (Attempt any two)

2(a) Central moments for the grouped data

Classes (marks): $0\!-\!10,10\!-\!20,20\!-\!30,30\!-\!40,40\!-\!50,50\!-\!60$
Frequencies: $1,6,10,15,11,7$. Midpoints: $5,15,25,35,45,55$.
Total $N=50$. Mean

$$\bar x=\frac{\sum f x}{N}=\frac{1\cdot5+6\cdot15+10\cdot25+15\cdot35+11\cdot45+7\cdot55}{50}=35.$$

Hence the central moments $\mu_r=\frac1N\sum f(x-\bar x)^r$:

$$\mu_1=0,\qquad \mu_2=164,\qquad \mu_3=-360,\qquad \mu_4=62000.$$

(These come from the midpoints; show a short working table if space allows.)

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2(b) Mean and s.d. for $dp=\dfrac{x^{\,n-1}\,dx}{B(n,m)(1+x)^{n+m}},\;x>0,\;n,m>0$

This is the Beta-prime (Type-VI) density with parameters $(n,m)$.

Using $\int_0^\infty \frac{x^{n+r-1}}{(1+x)^{n+m}}dx = B(n+r,m-r)$ (valid for $m>r$):

$$\mathbb E[X]=\frac{B(n+1,m-1)}{B(n,m)}=\frac{n}{m-1}\quad (m>1),$$ $$\mathbb E[X^2]=\frac{B(n+2,m-2)}{B(n,m)}=\frac{n(n+1)}{(m-1)(m-2)}\quad (m>2).$$

So

$$\operatorname{Var}(X)=\frac{n(n+m-1)}{(m-2)(m-1)^2},\qquad \text{s.d.}=\sqrt{\frac{n(n+m-1)}{(m-2)(m-1)^2}}.$$

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2(c) Chebyshev and the actual probability for $p(x)=2^{-x},\;x=1,2,\ldots$

This is geometric with parameter $p=\tfrac12$ (support starts at 1).
$\mu=\frac1p=2,\;\sigma^2=\frac{1-p}{p^2}=2$.

Chebyshev with $a=2$:

$$\Pr(|X-\mu|\le a)\ge 1-\frac{\sigma^2}{a^2} =1-\frac{2}{4}=\tfrac12.$$

Actual probability:

$$\Pr(|X-2|\le2)=\Pr(1\le X\le4)=\sum_{x=1}^4 2^{-x} =\frac12+\frac14+\frac18+\frac1{16}=\frac{15}{16}=0.9375.$$

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3. (Attempt any two)

3(a) Central Limit Theorem (Lindeberg–Levy form)

If $X_1,X_2,\ldots$ are i.i.d. with $\mathbb E[X_i]=\mu$ and $\operatorname{Var}(X_i)=\sigma^2\in(0,\infty)$, then

$$Z_n=\frac{\overline X_n-\mu}{\sigma/\sqrt n}\;\xRightarrow{d}\;N(0,1)\quad(n\to\infty).$$

Sketch proof: Consider the mgf of $Z_n$: (M_{Z_n}(t)=\big(M_{X_1}!\left(\tfrac{t}{\sqrt n}\right) e^{-t\mu/\sqrt n}\big)^n \to \exp!\left(\tfrac{t^2}{2}\right)) by a Taylor expansion of $M_{X_1}$ about 0. The limit is the mgf of $N(0,1)$, proving convergence in distribution.

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3(b) Least-squares parabola for points $(-1,2),(0,0),(0,1),(1,2)$

Fit $y=a+bx+cx^2$. Normal equations (or matrix least squares) give

$$\boxed{\,\hat y=0.5+1.5\,x^2\,}$$

($a=0.5,\;b=0,\;c=1.5$). Residual sum of squares $=0.5$.

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3(c) MGF of the normal and link with characteristic function

If $X\sim N(\mu,\sigma^2)$ with density $\frac{1}{\sqrt{2\pi}\sigma}\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$, then completing the square gives

$$M_X(t)=\mathbb E[e^{tX}] =\exp\!\left(\mu t+\frac{\sigma^2t^2}{2}\right).$$

The characteristic function is $\phi_X(t)=M_X(it)=\exp\!\left(i\mu t-\frac{\sigma^2t^2}{2}\right)$.
Moments satisfy $\mathbb E[X^r]=M_X^{(r)}(0)=i^{-r}\phi_X^{(r)}(0)$.

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4. (Attempt any two)

4(a) Short notes

(i) Null hypothesis $H_0$: the default claim about a population parameter (e.g., “no effect,” $p_1=p_2$, $\mu=\mu_0$). We attempt to reject $H_0$ using a test statistic and a significance level $\alpha$.

(ii) Degrees of freedom (d.f.): number of independent pieces of information used to estimate a parameter or compute a statistic (e.g., for sample variance, d.f. $=n-1$; for χ² GOF with $k$ classes and $s$ estimated parameters, d.f. $=k-1-s$).

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4(b) Uniformity of aircraft accidents over the week (χ² test)

Observed: $14,16,8,12,11,9,14$; $N=84$.
Expected under uniformity: $E=84/7=12$ in each day.

$$\chi^2=\sum\frac{(O-E)^2}{E}=\frac{(14-12)^2}{12}+\cdots+\frac{(14-12)^2}{12} =4.167.$$

d.f. $=7-1=6$. Critical value (5%): $12.59$.
Since $4.167<12.59$ ⇒ fail to reject uniformity. Accidents appear uniformly distributed.

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4(c) Compare proportions of boys with the defect in cities A and B

A: $n_1=900,\ \hat p_1=0.20\Rightarrow x_1=180$.
B: $n_2=1600,\ \hat p_2=0.185\Rightarrow x_2=296$.
Test $H_0:p_1=p_2$ vs two-sided.

Pooled $\hat p=\frac{x_1+x_2}{n_1+n_2}=0.1904$.
SE $=\sqrt{\hat p(1-\hat p)\left(\frac1{n_1}+\frac1{n_2}\right)}=0.01636$.

$$z=\frac{\hat p_1-\hat p_2}{\text{SE}}=\frac{0.015}{0.01636}=0.917.$$

At 5%, $|z|<1.96$ ⇒ not significant. No evidence of a real difference.

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5. (Attempt any two)

5(a) One-sample $t$ test for the mean

$n=16$, $\bar x=53$, $\sum (x_i-\bar x)^2=150\Rightarrow s^2=\frac{150}{15}=10$.
$H_0:\mu=56$ vs two-sided.

$$t=\frac{\bar x-\mu_0}{s/\sqrt n}=\frac{53-56}{\sqrt{10}/4}=-3.795.$$

d.f. $=15$; critical (5%) $t_{0.025,15}=2.131$.
$|t|>2.131\Rightarrow$ reject $H_0$. The sample is not from a population with mean 56.

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5(b) Student $t$ distribution: features, pdf, normal limit

For d.f. $\nu>0$, the pdf is

$$f_\nu(t)=\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu\pi}\,\Gamma\!\left(\frac{\nu}{2}\right)} \left(1+\frac{t^2}{\nu}\right)^{-(\nu+1)/2},\quad -\infty<t<\infty.$$

Main characteristics: symmetric about 0; heavier tails than normal;
$\mathbb E[T]=0$ for $\nu>1$; $\operatorname{Var}(T)=\frac{\nu}{\nu-2}$ for $\nu>2$; median 0; unimodal.
Normal limit: as $\nu\to\infty$,

$$\left(1+\frac{t^2}{\nu}\right)^{-(\nu+1)/2}\to e^{-t^2/2},$$

and the constant in front tends to $(2\pi)^{-1/2}$, so $t_\nu \Rightarrow N(0,1)$.

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5(c) Mean and variance of the $F$ distribution

If $F\sim F(\nu_1,\nu_2)$ with $\nu_1,\nu_2>0$, then

$$\mathbb E[F]=\frac{\nu_2}{\nu_2-2}\quad(\nu_2>2),$$ $$\operatorname{Var}(F)=\frac{2\nu_2^2(\nu_1+\nu_2-2)}{\nu_1(\nu_2-2)^2(\nu_2-4)} \quad(\nu_2>4).$$

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All parts solved — if you need any of these rewritten in your preferred step-by-step working format (tables, intermediate sums, or boxed final statements), say the word and I’ll format to match your exam style.