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Section A: Long Answer Questions

Attempt any TWO questions.

3 questions·10 marks each
1long10 marks

Define a probability distribution. Explain the binomial distribution with its mean and variance, and state the conditions under which it is applied.

Probability Distribution

A probability distribution is a mathematical function (or table) that assigns to each possible value of a random variable the probability of its occurrence. For a discrete random variable XX taking values x1,x2,x_1, x_2, \dots it is described by the probability mass function P(X=xi)=piP(X = x_i) = p_i with pi0p_i \ge 0 and ipi=1\sum_i p_i = 1; for a continuous variable it is described by a probability density function f(x)f(x) with f(x)0f(x) \ge 0 and f(x)dx=1\int_{-\infty}^{\infty} f(x)\,dx = 1.

Binomial Distribution

A discrete random variable XX follows a binomial distribution if it counts the number of successes in nn independent Bernoulli trials, each having success probability pp (and failure q=1pq = 1 - p). Its probability mass function is

P(X=x)=(nx)pxqnx,x=0,1,2,,n.P(X = x) = \binom{n}{x} p^{x} q^{\,n-x}, \qquad x = 0, 1, 2, \dots, n.

Mean and Variance

Mean μ=E(X)=np,Variance σ2=npq.\text{Mean } \mu = E(X) = np, \qquad \text{Variance } \sigma^{2} = npq.

Since q<1q < 1, the variance npqnpq is always less than the mean npnp, i.e. for the binomial distribution mean >> variance.

Outline of the mean: writing X=i=1nXiX = \sum_{i=1}^{n} X_i where each XiX_i is Bernoulli with E(Xi)=pE(X_i)=p, by linearity E(X)=E(Xi)=npE(X)=\sum E(X_i)=np; similarly Var(Xi)=pq\operatorname{Var}(X_i)=pq and by independence Var(X)=npq\operatorname{Var}(X)=npq.

Conditions for Application

The binomial distribution applies when:

  1. The number of trials nn is fixed and finite.
  2. Each trial has only two outcomes — success or failure (dichotomous).
  3. The trials are independent of one another.
  4. The probability of success pp remains constant from trial to trial.

Examples: number of heads in 10 tosses of a coin, number of defective items in a sample of fixed size, number of correct answers in a multiple-choice test.

probabilitydistribution
2long10 marks

Explain the normal distribution and its properties. The mean of a normal distribution is 50 and standard deviation is 10; find the probability that a value lies between 45 and 62.

Normal Distribution

A continuous random variable XX follows a normal distribution with mean μ\mu and standard deviation σ\sigma if its probability density function is

f(x)=1σ2πe(xμ)22σ2,<x<.f(x) = \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x-\mu)^2}{2\sigma^2}}, \qquad -\infty < x < \infty.

We write XN(μ,σ2)X \sim N(\mu, \sigma^2).

Properties

  1. The curve is bell-shaped and symmetric about the mean μ\mu.
  2. Mean = Median = Mode = μ\mu.
  3. It is unimodal; the maximum of f(x)f(x) occurs at x=μx = \mu.
  4. The curve is asymptotic to the xx-axis on both sides.
  5. Total area under the curve is 1, with half on each side of μ\mu.
  6. The points of inflection occur at x=μ±σx = \mu \pm \sigma.
  7. Empirical rule: about 68.27%, 95.45% and 99.73% of values lie within μ±σ\mu\pm\sigma, μ±2σ\mu\pm 2\sigma and μ±3σ\mu\pm 3\sigma respectively.
  8. Quartile deviation =0.6745σ= 0.6745\sigma and mean deviation =0.7979σ= 0.7979\sigma.
  9. The standard normal variate is Z=XμσN(0,1)Z = \dfrac{X - \mu}{\sigma} \sim N(0,1).

Numerical: P(45<X<62)P(45 < X < 62)

Given μ=50\mu = 50, σ=10\sigma = 10. Convert to ZZ-scores:

Z1=455010=0.5,Z2=625010=1.2.Z_1 = \frac{45 - 50}{10} = -0.5, \qquad Z_2 = \frac{62 - 50}{10} = 1.2.

Using the standard normal table (Φ\Phi = area from 00 to zz):

P(0<Z<0.5)=0.1915,P(0<Z<1.2)=0.3849.P(0 < Z < 0.5) = 0.1915, \qquad P(0 < Z < 1.2) = 0.3849.

Since the two points lie on opposite sides of the mean,

P(45<X<62)=P(0.5<Z<1.2)=0.1915+0.3849=0.5764.P(45 < X < 62) = P(-0.5 < Z < 1.2) = 0.1915 + 0.3849 = \boxed{0.5764}.

Hence the required probability is about 0.5764 (57.64%).

normal-distribution
3long10 marks

What is hypothesis testing? Explain the procedure of testing of hypothesis including null and alternative hypotheses, level of significance, types of errors and the critical region.

Hypothesis Testing

Hypothesis testing is a statistical procedure used to decide, on the basis of sample data, whether to accept or reject an assumption (hypothesis) made about a population parameter. It quantifies how strongly the sample evidence contradicts the assumption.

Key Concepts

1. Null Hypothesis (H0H_0): A statement of no effect or no difference that is assumed true until evidence suggests otherwise, e.g. H0:μ=μ0H_0: \mu = \mu_0.

2. Alternative Hypothesis (H1H_1): The statement accepted if H0H_0 is rejected. It may be:

  • Two-tailed: H1:μμ0H_1: \mu \ne \mu_0
  • Right-tailed: H1:μ>μ0H_1: \mu > \mu_0
  • Left-tailed: H1:μ<μ0H_1: \mu < \mu_0

3. Level of Significance (α\alpha): The maximum probability of rejecting H0H_0 when it is actually true (probability of a Type I error). Common values are 0.050.05 (5%) and 0.010.01 (1%).

4. Types of Errors:

H0H_0 TrueH0H_0 False
Reject H0H_0Type I error (α\alpha)Correct decision
Accept H0H_0Correct decisionType II error (β\beta)
  • Type I error (α\alpha): rejecting a true H0H_0.
  • Type II error (β\beta): accepting a false H0H_0. The quantity 1β1 - \beta is the power of the test.

5. Critical Region (Rejection Region): The set of values of the test statistic for which H0H_0 is rejected. Its area equals α\alpha. The boundary value(s) separating it from the acceptance region are the critical value(s).

Procedure (Steps)

  1. Set up the null hypothesis H0H_0 and alternative hypothesis H1H_1.
  2. Choose the level of significance α\alpha.
  3. Select an appropriate test statistic (Z, t, χ2\chi^2, F) based on sample size and what is known.
  4. Determine the critical value and the critical (rejection) region from statistical tables.
  5. Compute the value of the test statistic from the sample data.
  6. Decision: If the computed value falls in the critical region, reject H0H_0; otherwise accept (fail to reject) H0H_0.
  7. State the conclusion in the context of the problem.
hypothesis-testing
B

Section B: Short Answer Questions

Attempt any EIGHT questions.

9 questions·5 marks each
4short5 marks

State and explain the addition and multiplication theorems of probability with examples.

Addition Theorem of Probability

For any two events AA and BB,

P(AB)=P(A)+P(B)P(AB).P(A \cup B) = P(A) + P(B) - P(A \cap B).

If AA and BB are mutually exclusive (AB=A \cap B = \varnothing), this reduces to

P(AB)=P(A)+P(B).P(A \cup B) = P(A) + P(B).

Example: Drawing one card from a pack, P(King or Queen)=452+452=852=213P(\text{King or Queen}) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13} (mutually exclusive). For P(King or Heart)=452+1352152=1652=413P(\text{King or Heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} (not mutually exclusive).

Multiplication Theorem of Probability

For any two events AA and BB,

P(AB)=P(A)P(BA)=P(B)P(AB),P(A \cap B) = P(A)\,P(B \mid A) = P(B)\,P(A \mid B),

where P(BA)P(B \mid A) is the conditional probability of BB given AA. If AA and BB are independent, then P(BA)=P(B)P(B\mid A) = P(B) and

P(AB)=P(A)P(B).P(A \cap B) = P(A)\,P(B).

Example: Two cards drawn one after another without replacement, P(both Kings)=452×351=1221P(\text{both Kings}) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{221}. Tossing two coins (independent), P(both heads)=12×12=14P(\text{both heads}) = \frac{1}{2}\times\frac{1}{2} = \frac{1}{4}.

probability
5short5 marks

Explain the Poisson distribution with its mean and variance and state its applications.

Poisson Distribution

The Poisson distribution is a discrete distribution that models the number of times a rare event occurs in a fixed interval of time, space or volume when occurrences are independent and at a constant average rate. A random variable XX follows a Poisson distribution with parameter λ\lambda (the mean number of occurrences) if

P(X=x)=eλλxx!,x=0,1,2,;  λ>0.P(X = x) = \frac{e^{-\lambda}\,\lambda^{x}}{x!}, \qquad x = 0, 1, 2, \dots; \; \lambda > 0.

It is the limiting form of the binomial distribution when nn \to \infty, p0p \to 0, with np=λnp = \lambda finite.

Mean and Variance

Mean=E(X)=λ,Variance=Var(X)=λ.\text{Mean} = E(X) = \lambda, \qquad \text{Variance} = \operatorname{Var}(X) = \lambda.

A characteristic feature is that the mean equals the variance (μ=σ2=λ)(\mu = \sigma^2 = \lambda).

Applications

Used for counting rare, independent events such as:

  1. Number of telephone calls arriving at an exchange per minute.
  2. Number of printing/typing errors per page of a book.
  3. Number of accidents on a highway per day.
  4. Number of defective items in a large batch.
  5. Number of customers (or network packets/requests) arriving at a server per unit time.
  6. Number of radioactive particles emitted per second.
poisson
6short5 marks

Define a random variable. Differentiate between discrete and continuous random variables with examples.

Random Variable

A random variable is a real-valued function that assigns a numerical value to each outcome (sample point) of a random experiment. It is usually denoted by a capital letter XX, and a particular value by a small letter xx. For example, if a coin is tossed twice and XX denotes the number of heads, then XX takes the values 0,1,20, 1, 2.

Discrete vs Continuous Random Variables

BasisDiscrete Random VariableContinuous Random Variable
ValuesTakes a finite or countably infinite set of isolated valuesTakes any value within an interval (uncountable)
Described byProbability mass function P(X=x)=p(x)P(X=x)=p(x)Probability density function f(x)f(x)
Probability of a pointP(X=x)P(X=x) can be non-zeroP(X=x)=0P(X=x)=0; only P(a<X<b)P(a<X<b) is meaningful
Total probabilityxp(x)=1\sum_x p(x) = 1f(x)dx=1\int_{-\infty}^{\infty} f(x)\,dx = 1
ExampleNumber of heads in 3 tosses; number of defective bulbsHeight, weight, temperature, time of a person/object

Examples

  • Discrete: number of students present in a class, number of cars passing a point in an hour.
  • Continuous: the exact weight of a person, the lifetime of an electric bulb, daily rainfall.
random-variable
7short5 marks

Define mathematical expectation. State and prove its properties.

Mathematical Expectation

The mathematical expectation (or expected value) of a random variable is the long-run average value it takes, weighted by probabilities. For a discrete random variable XX with pmf p(x)p(x),

E(X)=xxp(x),E(X) = \sum_x x\,p(x),

and for a continuous random variable with pdf f(x)f(x),

E(X)=xf(x)dx.E(X) = \int_{-\infty}^{\infty} x\,f(x)\,dx.

Properties (with proofs)

1. Expectation of a constant. If kk is a constant, E(k)=kE(k) = k.

Proof: E(k)=xkp(x)=kxp(x)=k1=k.E(k) = \sum_x k\,p(x) = k\sum_x p(x) = k\cdot 1 = k.

2. Multiplication by a constant. E(kX)=kE(X)E(kX) = k\,E(X).

Proof: E(kX)=xkxp(x)=kxxp(x)=kE(X).E(kX) = \sum_x kx\,p(x) = k\sum_x x\,p(x) = k\,E(X).

3. Addition (linearity). E(X+Y)=E(X)+E(Y)E(X + Y) = E(X) + E(Y) for any random variables.

Proof (discrete): E(X+Y)=xy(x+y)p(x,y)=xyxp(x,y)+xyyp(x,y)=E(X)+E(Y).E(X+Y) = \sum_x\sum_y (x+y)p(x,y) = \sum_x\sum_y x\,p(x,y) + \sum_x\sum_y y\,p(x,y) = E(X) + E(Y).

Combining 2 and 3: E(aX+b)=aE(X)+bE(aX + b) = aE(X) + b.

4. Product for independent variables. If XX and YY are independent, E(XY)=E(X)E(Y)E(XY) = E(X)\,E(Y).

Proof: For independent variables p(x,y)=p(x)p(y)p(x,y) = p(x)p(y), so E(XY)=xyxyp(x)p(y)=(xxp(x))(yyp(y))=E(X)E(Y).E(XY) = \sum_x\sum_y xy\,p(x)p(y) = \left(\sum_x x\,p(x)\right)\left(\sum_y y\,p(y)\right) = E(X)E(Y).

expectation
8short5 marks

Explain the t-test for testing the significance of the difference between two sample means.

t-test for Difference of Two Sample Means

This test checks whether two independent small samples (sizes n1,n2<30n_1, n_2 < 30) drawn from normal populations with equal (unknown) variances have significantly different means.

Hypotheses: H0:μ1=μ2H_0: \mu_1 = \mu_2 (no difference) against H1:μ1μ2H_1: \mu_1 \ne \mu_2 (two-tailed) or one-sided alternative.

Test statistic:

t=xˉ1xˉ2S1n1+1n2,t = \frac{\bar{x}_1 - \bar{x}_2}{S\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}},

where the pooled standard deviation is

S2=(x1xˉ1)2+(x2xˉ2)2n1+n22=n1s12+n2s22n1+n22.S^2 = \frac{\sum (x_1 - \bar{x}_1)^2 + \sum (x_2 - \bar{x}_2)^2}{n_1 + n_2 - 2} = \frac{n_1 s_1^2 + n_2 s_2^2}{n_1 + n_2 - 2}.

The statistic follows the t-distribution with ν=n1+n22\nu = n_1 + n_2 - 2 degrees of freedom.

Decision rule: Compute t|t| and compare with the table value tα,νt_{\alpha,\nu} at significance level α\alpha.

  • If ttα,ν|t| \le t_{\alpha,\nu}: accept H0H_0 — the difference is not significant.
  • If t>tα,ν|t| > t_{\alpha,\nu}: reject H0H_0 — the difference is significant.

Assumptions: the parent populations are normal, the two samples are independent, and the population variances are equal (homogeneous).

t-test
9short5 marks

Explain the z-test for a large sample test of a single mean with an example.

z-test for a Single Mean (Large Sample)

When the sample size is large (n30n \ge 30), the sampling distribution of the mean is approximately normal, so a z-test is used to test whether a sample mean xˉ\bar{x} differs significantly from a hypothesised population mean μ\mu.

Hypotheses: H0:μ=μ0H_0: \mu = \mu_0 versus H1:μμ0H_1: \mu \ne \mu_0 (two-tailed).

Test statistic:

Z=xˉμ0σ/n,Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}},

where σ\sigma is the population standard deviation (if unknown, the sample s.d. ss is used since nn is large). ZZ follows the standard normal distribution.

Decision rule (5% level): reject H0H_0 if Z>1.96|Z| > 1.96 (for 1%, if Z>2.58|Z| > 2.58); otherwise accept H0H_0.

Example

A sample of n=64n = 64 items has mean xˉ=52\bar{x} = 52, drawn from a population with mean μ0=50\mu_0 = 50 and σ=8\sigma = 8. Test at 5% whether the sample mean differs from 50.

Z=52508/64=28/8=21=2.0.Z = \frac{52 - 50}{8/\sqrt{64}} = \frac{2}{8/8} = \frac{2}{1} = 2.0.

Since Z=2.0>1.96|Z| = 2.0 > 1.96, we reject H0H_0: the sample mean differs significantly from 50 at the 5% level of significance.

z-test
10short5 marks

Define Karl Pearson's coefficient of correlation and state its properties.

Karl Pearson's Coefficient of Correlation

Correlation measures the degree and direction of the linear relationship between two variables XX and YY. Karl Pearson's coefficient of correlation (product-moment correlation), denoted rr, is defined as

r=Cov(X,Y)σXσY=(xxˉ)(yyˉ)(xxˉ)2(yyˉ)2.r = \frac{\operatorname{Cov}(X,Y)}{\sigma_X\,\sigma_Y} = \frac{\sum (x-\bar{x})(y-\bar{y})}{\sqrt{\sum (x-\bar{x})^2}\,\sqrt{\sum (y-\bar{y})^2}}.

An equivalent computational form is

r=nxyxynx2(x)2ny2(y)2.r = \frac{n\sum xy - \sum x \sum y}{\sqrt{n\sum x^2 - (\sum x)^2}\,\sqrt{n\sum y^2 - (\sum y)^2}}.

Properties

  1. Range: rr always lies between 1-1 and +1+1, i.e. 1r+1-1 \le r \le +1.
  2. Interpretation: r=+1r = +1 perfect positive, r=1r = -1 perfect negative, r=0r = 0 no linear correlation.
  3. Independent of origin and scale (units): rr is unchanged if each value is shifted or multiplied by a constant; it is a pure number.
  4. Symmetric: rXY=rYXr_{XY} = r_{YX}.
  5. It is the geometric mean of the two regression coefficients: r=±bxybyxr = \pm\sqrt{b_{xy}\,b_{yx}}, taking the sign of the regression coefficients.
  6. If XX and YY are independent then r=0r = 0 (the converse is not necessarily true).
correlation
11short5 marks

What are regression coefficients? State their properties.

Regression Coefficients

In linear regression between two variables XX and YY, the regression coefficient is the slope of the line of regression — it measures the average change in the dependent variable per unit change in the independent variable.

  • Regression coefficient of YY on XX:
byx=rσyσx=Cov(X,Y)σx2.b_{yx} = r\,\frac{\sigma_y}{\sigma_x} = \frac{\operatorname{Cov}(X,Y)}{\sigma_x^2}.
  • Regression coefficient of XX on YY:
bxy=rσxσy=Cov(X,Y)σy2.b_{xy} = r\,\frac{\sigma_x}{\sigma_y} = \frac{\operatorname{Cov}(X,Y)}{\sigma_y^2}.

Properties

  1. Correlation as geometric mean: the correlation coefficient is the geometric mean of the two regression coefficients,
r=±bxybyx.r = \pm\sqrt{b_{xy}\cdot b_{yx}}.
  1. Same sign: both regression coefficients have the same sign, which is also the sign of rr. If one is positive both are positive, and so is rr.
  2. Product 1\le 1: since 1r1-1 \le r \le 1,   bxybyx=r21\;b_{xy}\cdot b_{yx} = r^2 \le 1. Hence both coefficients cannot exceed unity simultaneously; if one is greater than 1, the other must be less than 1.
  3. Independent of change of origin but not of scale.
  4. The arithmetic mean of the two regression coefficients is greater than or equal to rr, i.e. bxy+byx2r\dfrac{b_{xy}+b_{yx}}{2} \ge r.
  5. The two regression lines intersect at the point (xˉ,yˉ)(\bar{x}, \bar{y}).
regression
12short5 marks

Explain the concept of sampling distribution and standard error.

Sampling Distribution

When all possible samples of a fixed size nn are drawn from a population, a statistic (such as the sample mean xˉ\bar{x}, proportion pp, or variance s2s^2) varies from sample to sample. The probability distribution of all possible values of that statistic is called its sampling distribution.

For example, the sampling distribution of the mean lists every possible xˉ\bar{x} with its probability. By the Central Limit Theorem, for large nn the sampling distribution of xˉ\bar{x} is approximately normal with mean μ\mu and standard deviation σ/n\sigma/\sqrt{n}, regardless of the shape of the parent population.

Standard Error (S.E.)

The standard error of a statistic is the standard deviation of its sampling distribution. It measures the variability of the statistic due to sampling and indicates how precisely the sample statistic estimates the population parameter.

Common standard errors:

  • Mean:   S.E.(xˉ)=σn\;\text{S.E.}(\bar{x}) = \dfrac{\sigma}{\sqrt{n}}
  • Proportion:   S.E.(p)=PQn\;\text{S.E.}(p) = \sqrt{\dfrac{PQ}{n}}, where Q=1PQ = 1-P.

Uses: the standard error is used (i) to test the significance of a statistic (test statistic = (estimate − parameter)/S.E.), and (ii) to construct confidence intervals. A smaller standard error (larger nn) means a more reliable estimate.

samplingdistribution

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