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A

Section A: Long Answer Questions

Attempt any TWO questions.

3 questions·10 marks each
1long10 marks

Explain the concept of a measure of dispersion. Calculate the quartile deviation, mean deviation and standard deviation for the given data and compare them.

Measure of Dispersion

A measure of dispersion (or variation) describes how much the individual observations in a data set are spread out or scattered about a central value (mean, median). While a measure of central tendency gives a single representative value, a measure of dispersion tells us how reliable that value is. Small dispersion means the data are homogeneous and clustered; large dispersion means they are heterogeneous and widely scattered.

Common measures: Range, Quartile Deviation (Q.D.), Mean Deviation (M.D.) and Standard Deviation (S.D.). Range and Q.D. are positional (based on quartiles); M.D. and S.D. are calculation-based (based on deviations from an average).

Worked Example

Take the data: 5,7,9,11,13,15,175, 7, 9, 11, 13, 15, 17 (n = 7), already arranged in order.

Mean xˉ=5+7+9+11+13+15+177=777=11\bar{x} = \dfrac{5+7+9+11+13+15+17}{7} = \dfrac{77}{7} = 11.

(a) Quartile Deviation

Position of Q1=(n+14)th=2ndQ_1 = \left(\dfrac{n+1}{4}\right)^{th} = 2^{nd} item =7= 7.

Position of Q3=(3(n+1)4)th=6thQ_3 = \left(\dfrac{3(n+1)}{4}\right)^{th} = 6^{th} item =15= 15.

Q.D.=Q3Q12=1572=4Q.D. = \frac{Q_3 - Q_1}{2} = \frac{15 - 7}{2} = 4

(b) Mean Deviation (about the mean)

M.D.=xxˉnM.D. = \frac{\sum |x - \bar{x}|}{n}
xx57911131517
$x-11$64202

xxˉ=24\sum|x-\bar{x}| = 24, so M.D.=247=3.43M.D. = \dfrac{24}{7} = 3.43.

(c) Standard Deviation

S.D.=(xxˉ)2nS.D. = \sqrt{\frac{\sum (x-\bar{x})^2}{n}}

(xxˉ)2=36+16+4+0+4+16+36=112\sum(x-\bar{x})^2 = 36+16+4+0+4+16+36 = 112.

S.D.=1127=16=4S.D. = \sqrt{\frac{112}{7}} = \sqrt{16} = 4

Comparison

MeasureValueBasis
Quartile Deviation4.00Middle 50% of data only
Mean Deviation3.43Absolute deviations from mean
Standard Deviation4.00Squared deviations from mean

For this symmetric data the values are close. In general, S.D. is the most reliable measure because it uses every observation, is amenable to algebraic treatment and is the basis of further statistical analysis; Q.D. is preferred for open-ended distributions, and M.D. is simple but less suitable for further computation (theoretically M.D.<S.D.M.D. < S.D. for most distributions).

dispersion
2long10 marks

Define conditional probability and independent events. State Bayes' theorem and apply it to solve a problem on conditional probability.

Conditional Probability

The conditional probability of event AA given that event BB has already occurred is

P(AB)=P(AB)P(B),P(B)>0.P(A\mid B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0.

It re-scales the sample space to those outcomes in which BB occurs.

Independent Events

Two events AA and BB are independent if the occurrence of one does not affect the probability of the other:

P(AB)=P(A)P(B),equivalentlyP(AB)=P(A).P(A \cap B) = P(A)\,P(B), \quad\text{equivalently}\quad P(A\mid B) = P(A).

Bayes' Theorem

If B1,B2,,BnB_1, B_2, \dots, B_n are mutually exclusive and exhaustive events with P(Bi)>0P(B_i) > 0, and AA is any event with P(A)>0P(A) > 0, then

P(BiA)=P(Bi)P(ABi)j=1nP(Bj)P(ABj).P(B_i \mid A) = \frac{P(B_i)\,P(A\mid B_i)}{\sum_{j=1}^{n} P(B_j)\,P(A\mid B_j)}.

The P(Bi)P(B_i) are prior probabilities and P(BiA)P(B_i\mid A) are posterior probabilities; the denominator is P(A)P(A) by the law of total probability.

Applied Problem

Three machines B1,B2,B3B_1, B_2, B_3 produce 50%, 30% and 20% of a factory's output, with defective rates 3%, 4% and 5% respectively. An item is found defective; what is the probability it came from machine B3B_3?

Priors: P(B1)=0.5, P(B2)=0.3, P(B3)=0.2P(B_1)=0.5,\ P(B_2)=0.3,\ P(B_3)=0.2. Likelihoods (defective =A=A): P(AB1)=0.03, P(AB2)=0.04, P(AB3)=0.05P(A\mid B_1)=0.03,\ P(A\mid B_2)=0.04,\ P(A\mid B_3)=0.05.

Total probability of a defective item:

P(A)=(0.5)(0.03)+(0.3)(0.04)+(0.2)(0.05)=0.015+0.012+0.010=0.037.P(A) = (0.5)(0.03) + (0.3)(0.04) + (0.2)(0.05) = 0.015 + 0.012 + 0.010 = 0.037.

By Bayes' theorem:

P(B3A)=(0.2)(0.05)0.037=0.0100.037=0.270.P(B_3 \mid A) = \frac{(0.2)(0.05)}{0.037} = \frac{0.010}{0.037} = 0.270.

Conclusion: Given the item is defective, there is about a 27% chance it was produced by machine B3B_3.

probabilitybayes
3long10 marks

Define mathematical expectation. Find the mean and variance of a discrete random variable from its probability distribution. State the properties of expectation.

Mathematical Expectation

The mathematical expectation (expected value) of a random variable is its long-run average value, weighting each value by its probability. For a discrete random variable XX taking values xix_i with probabilities pi=P(X=xi)p_i = P(X=x_i),

E(X)=μ=ixipi,where ipi=1.E(X) = \mu = \sum_i x_i\, p_i, \qquad \text{where } \sum_i p_i = 1.

Mean and Variance from a Distribution

Consider the distribution:

xx0123
P(x)P(x)0.10.30.40.2

Mean:

E(X)=xP(x)=0(0.1)+1(0.3)+2(0.4)+3(0.2)=0+0.3+0.8+0.6=1.7.E(X) = \sum x P(x) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0 + 0.3 + 0.8 + 0.6 = 1.7.

For variance compute E(X2)E(X^2):

E(X2)=x2P(x)=0(0.1)+1(0.3)+4(0.4)+9(0.2)=0+0.3+1.6+1.8=3.7.E(X^2) = \sum x^2 P(x) = 0(0.1) + 1(0.3) + 4(0.4) + 9(0.2) = 0 + 0.3 + 1.6 + 1.8 = 3.7. Var(X)=E(X2)[E(X)]2=3.7(1.7)2=3.72.89=0.81.Var(X) = E(X^2) - [E(X)]^2 = 3.7 - (1.7)^2 = 3.7 - 2.89 = 0.81.

Standard deviation σ=0.81=0.9\sigma = \sqrt{0.81} = 0.9.

Properties of Expectation

  1. E(c)=cE(c) = c for any constant cc.
  2. E(cX)=cE(X)E(cX) = c\,E(X).
  3. E(aX+b)=aE(X)+bE(aX + b) = a\,E(X) + b (linearity).
  4. E(X+Y)=E(X)+E(Y)E(X + Y) = E(X) + E(Y) — addition theorem (always holds).
  5. If XX and YY are independent, E(XY)=E(X)E(Y)E(XY) = E(X)\,E(Y) — multiplication theorem.
  6. Var(X)=E(X2)[E(X)]2Var(X) = E(X^2) - [E(X)]^2, and Var(aX+b)=a2Var(X)Var(aX+b) = a^2 Var(X).
expectationrandom-variable
B

Section B: Short Answer Questions

Attempt any EIGHT questions.

9 questions·5 marks each
4short5 marks

Define mathematical expectation.

The mathematical expectation (or expected value) of a random variable XX is its probability-weighted average value, denoted E(X)E(X) or μ\mu.

  • Discrete: E(X)=ixipiE(X) = \sum_i x_i\, p_i, where pi=P(X=xi)p_i = P(X = x_i) and pi=1\sum p_i = 1.
  • Continuous: E(X)=xf(x)dxE(X) = \displaystyle\int_{-\infty}^{\infty} x\, f(x)\, dx, where f(x)f(x) is the probability density function.

It represents the long-run average value of XX over many repetitions of the experiment and serves as the theoretical mean of the distribution.

expectation
5short5 marks

State Bayes' theorem.

Bayes' theorem relates posterior probability to prior probability and likelihood. If B1,B2,,BnB_1, B_2, \dots, B_n are mutually exclusive and exhaustive events (a partition of the sample space) with P(Bi)>0P(B_i) > 0, and AA is any event with P(A)>0P(A) > 0, then

P(BiA)=P(Bi)P(ABi)j=1nP(Bj)P(ABj).P(B_i \mid A) = \frac{P(B_i)\,P(A \mid B_i)}{\sum_{j=1}^{n} P(B_j)\,P(A \mid B_j)}.

Here P(Bi)P(B_i) are the prior probabilities, P(ABi)P(A\mid B_i) the likelihoods, and P(BiA)P(B_i\mid A) the posterior probabilities. The denominator equals P(A)P(A) by the law of total probability.

bayes
6short5 marks

What is the difference between a parameter and a statistic?

A parameter is a numerical characteristic of the entire population, whereas a statistic is a numerical characteristic computed from a sample.

AspectParameterStatistic
Computed fromWhole populationA sample drawn from the population
ValueFixed (usually unknown) constantVaries from sample to sample (random)
Notationμ\mu (mean), σ\sigma (S.D.), PP (proportion)xˉ\bar{x} (mean), ss (S.D.), pp (proportion)
PurposeThe quantity we want to knowUsed to estimate the parameter

In short, a statistic (e.g. sample mean xˉ\bar{x}) is used to estimate the corresponding unknown parameter (e.g. population mean μ\mu).

basics
7short5 marks

Define the variance of a random variable.

The variance of a random variable XX measures the average squared deviation of XX from its mean μ=E(X)\mu = E(X). It quantifies the spread of the distribution:

Var(X)=σ2=E[(Xμ)2]=E(X2)[E(X)]2.Var(X) = \sigma^2 = E\big[(X - \mu)^2\big] = E(X^2) - [E(X)]^2.
  • Discrete: Var(X)=i(xiμ)2piVar(X) = \sum_i (x_i - \mu)^2 p_i.
  • Continuous: Var(X)=(xμ)2f(x)dxVar(X) = \displaystyle\int_{-\infty}^{\infty} (x-\mu)^2 f(x)\, dx.

Variance is always non-negative, and its positive square root σ=Var(X)\sigma = \sqrt{Var(X)} is the standard deviation. Also Var(aX+b)=a2Var(X)Var(aX+b) = a^2\,Var(X).

random-variable
8short5 marks

What is a probability distribution?

A probability distribution of a random variable is a description that assigns probabilities to all the possible values the variable can take, listing each value together with its probability of occurrence.

  • Discrete distribution: specified by a probability mass function P(X=xi)=piP(X=x_i)=p_i satisfying pi0p_i \ge 0 and ipi=1\sum_i p_i = 1.
  • Continuous distribution: specified by a probability density function f(x)f(x) satisfying f(x)0f(x) \ge 0 and f(x)dx=1\int_{-\infty}^{\infty} f(x)\,dx = 1.

Example (discrete): For a fair die, P(X=x)=16P(X=x) = \frac{1}{6} for x=1,2,,6x = 1,2,\dots,6. Examples include the Binomial, Poisson and Normal distributions.

distribution
9short5 marks

Define the weighted arithmetic mean.

The weighted arithmetic mean is an average in which each value xix_i is given a weight wiw_i reflecting its relative importance, instead of treating all observations equally. It is defined as

xˉw=i=1nwixii=1nwi.\bar{x}_w = \frac{\sum_{i=1}^{n} w_i x_i}{\sum_{i=1}^{n} w_i}.

It is used when the items do not contribute equally (e.g. computing a GPA where credit hours are weights, or an index number). When all weights are equal it reduces to the ordinary arithmetic mean.

Example: Marks 80, 70, 90 with weights (credits) 3, 2, 5: xˉw=3(80)+2(70)+5(90)3+2+5=83010=83\bar{x}_w = \dfrac{3(80)+2(70)+5(90)}{3+2+5} = \dfrac{830}{10} = 83.

central-tendency
10short5 marks

State the properties of expectation.

Properties of mathematical expectation:

  1. Constant: E(c)=cE(c) = c for any constant cc.
  2. Scalar multiple: E(cX)=cE(X)E(cX) = c\,E(X).
  3. Linearity: E(aX+b)=aE(X)+bE(aX + b) = a\,E(X) + b.
  4. Addition theorem: E(X+Y)=E(X)+E(Y)E(X + Y) = E(X) + E(Y) (holds for any X,YX, Y).
  5. Multiplication theorem: if XX and YY are independent, E(XY)=E(X)E(Y)E(XY) = E(X)\,E(Y).
  6. E(X)E(X) lies between the minimum and maximum values of XX; and if X0X \ge 0 then E(X)0E(X) \ge 0.
  7. Relation to variance: Var(X)=E(X2)[E(X)]2Var(X) = E(X^2) - [E(X)]^2.
expectation
11short5 marks

What is the interquartile range?

The interquartile range (IQR) is the difference between the third quartile Q3Q_3 (75th percentile) and the first quartile Q1Q_1 (25th percentile):

IQR=Q3Q1.IQR = Q_3 - Q_1.

It is a measure of dispersion covering the middle 50% of the data and is not affected by extreme values (outliers), making it a robust measure of spread. The related quartile deviation (semi-interquartile range) is Q3Q12\dfrac{Q_3 - Q_1}{2}.

Example: If Q1=25Q_1 = 25 and Q3=45Q_3 = 45, then IQR=4525=20IQR = 45 - 25 = 20.

dispersion
12short5 marks

Find the probability of getting a head when a fair coin is tossed once.

When a fair coin is tossed once, the sample space is S={H,T}S = \{H, T\} with n(S)=2n(S) = 2 equally likely outcomes. The favourable event A={H}A = \{H\} has n(A)=1n(A) = 1.

P(head)=n(A)n(S)=12=0.5.P(\text{head}) = \frac{n(A)}{n(S)} = \frac{1}{2} = 0.5.

Thus the probability of getting a head is 12\tfrac{1}{2} or 0.5 (50%).

probability

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