November 2025

Exam - Semester Examination
Session - November 2025

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Question 1 (20 Marks)

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Question Statement

Calculate standard deviation, variance and the coefficient of quartile deviation, coefficient of variation, Karl Pearson's coefficient of skewness for the following data. (20 Marks)

Sales	15	18	25	27	30	35
Expenditure	50	65	82	95	110	120

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Solution 1 (20 Marks)

Part A: Statistical Measures for Sales ($X$) (10 Marks)

Given ordered observations ($n = 6$): $X = [15, 18, 25, 27, 30, 35]$

1. Mean and Deviations

• Sum: $\sum X = 15 + 18 + 25 + 27 + 30 + 35 = 150$
• Mean: $\bar{X} = \frac{\sum X}{n} = \frac{150}{6} = \mathbf{25.0}$

$X_i$	$(X_i - \bar{X})$	$(X_i - \bar{X})^2$
15	$-10$	100
18	$-7$	49
25	$0$	0
27	$+2$	4
30	$+5$	25
35	$+10$	100
Total	$\sum = 0$	$\sum (X_i - \bar{X})^2 = \mathbf{278}$

2. Variance and Standard Deviation

• Population Variance ($\sigma^2$): $\sigma^2 = \frac{\sum (X_i - \bar{X})^2}{n} = \frac{278}{6} \approx \mathbf{46.3333}$ (Sample variance $s^2 = \frac{278}{5} = 55.60$)
• Population Standard Deviation ($\sigma$): $\sigma = \sqrt{46.3333} \approx \mathbf{6.8069}$ (Sample standard deviation $s = \sqrt{55.60} \approx 7.4565$)

3. Quartiles and Coefficient of Quartile Deviation

• Median ($Q_2$): For $n=6$, average of 3rd and 4th items: $M = \frac{25 + 27}{2} = 26.0$
• First Quartile ($Q_1$): Position $\frac{n+1}{4} = \frac{7}{4} = 1.75$: $Q_1 = X_{(1)} + 0.75(X_{(2)} - X_{(1)}) = 15 + 0.75(18 - 15) = 15 + 2.25 = \mathbf{17.25}$ (Using lower-half median method: $Q_1 = 18.0$)
• Third Quartile ($Q_3$): Position $\frac{3(n+1)}{4} = \frac{21}{4} = 5.25$: $Q_3 = X_{(5)} + 0.25(X_{(6)} - X_{(5)}) = 30 + 0.25(35 - 30) = 30 + 1.25 = \mathbf{31.25}$ (Using upper-half median method: $Q_3 = 30.0$)
• Quartile Deviation (Semi-Interquartile Range): $\text{QD} = \frac{Q_3 - Q_1}{2} = \frac{31.25 - 17.25}{2} = \frac{14.0}{2} = \mathbf{7.0}$
• Coefficient of Quartile Deviation: $\text{Coeff of QD} = \frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{31.25 - 17.25}{31.25 + 17.25} = \frac{14.0}{48.5} \approx \mathbf{0.2887}$

4. Coefficient of Variation (CV)

$\text{CV} = \frac{\sigma}{\bar{X}} \times 100 = \frac{6.8069}{25.0} \times 100 \approx \mathbf{27.23\%}$

5. Karl Pearson's Coefficient of Skewness ($Sk_p$)

Since all data values are distinct (no repeated mode), use the median formulation: $Sk_p = \frac{3(\text{Mean} - \text{Median})}{\sigma} = \frac{3(25.0 - 26.0)}{6.8069} = \frac{-3.0}{6.8069} \approx \mathbf{-0.4407}$

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Part B: Statistical Measures for Expenditure ($Y$) (10 Marks)

Given ordered observations ($n = 6$): $Y = [50, 65, 82, 95, 110, 120]$

1. Mean and Deviations

• Sum: $\sum Y = 50 + 65 + 82 + 95 + 110 + 120 = 522$
• Mean: $\bar{Y} = \frac{\sum Y}{n} = \frac{522}{6} = \mathbf{87.0}$

$Y_i$	$(Y_i - \bar{Y})$	$(Y_i - \bar{Y})^2$
50	$-37$	1369
65	$-22$	484
82	$-5$	25
95	$+8$	64
110	$+23$	529
120	$+33$	1089
Total	$\sum = 0$	$\sum (Y_i - \bar{Y})^2 = \mathbf{3560}$

2. Variance and Standard Deviation

• Population Variance ($\sigma^2$): $\sigma^2 = \frac{\sum (Y_i - \bar{Y})^2}{n} = \frac{3560}{6} \approx \mathbf{593.3333}$ (Sample variance $s^2 = \frac{3560}{5} = 712.0$)
• Population Standard Deviation ($\sigma$): $\sigma = \sqrt{593.3333} \approx \mathbf{24.3584}$ (Sample standard deviation $s = \sqrt{712.0} \approx 26.6833$)

3. Quartiles and Coefficient of Quartile Deviation

• Median ($Q_2$): $M = \frac{82 + 95}{2} = \mathbf{88.5}$
• First Quartile ($Q_1$): Position $1.75$: $Q_1 = 50 + 0.75(65 - 50) = 50 + 11.25 = \mathbf{61.25}$ (Lower-half median: $Q_1 = 65.0$)
• Third Quartile ($Q_3$): Position $5.25$: $Q_3 = 110 + 0.25(120 - 110) = 110 + 2.5 = \mathbf{112.50}$ (Upper-half median: $Q_3 = 110.0$)
• Quartile Deviation: $\text{QD} = \frac{Q_3 - Q_1}{2} = \frac{112.50 - 61.25}{2} = \frac{51.25}{2} = \mathbf{25.625}$
• Coefficient of Quartile Deviation: $\text{Coeff of QD} = \frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{51.25}{173.75} \approx \mathbf{0.2950}$

4. Coefficient of Variation (CV)

$\text{CV} = \frac{\sigma}{\bar{Y}} \times 100 = \frac{24.3584}{87.0} \times 100 \approx \mathbf{28.00\%}$

5. Karl Pearson's Coefficient of Skewness ($Sk_p$)

$Sk_p = \frac{3(\text{Mean} - \text{Median})}{\sigma} = \frac{3(87.0 - 88.5)}{24.3584} = \frac{-4.5}{24.3584} \approx \mathbf{-0.1847}$

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Examiner's Note

Both variables exhibit mild negative skewness ($Sk_p < 0$). In datasets where all values are distinct, always use Pearson's second empirical skewness formula $Sk_p = \frac{3(\text{Mean} - \text{Median})}{\sigma}$.

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Question 2 (20 Marks)

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Question Statement

• i) Discuss the Binomial and Poisson distribution, its properties, and its application in solving real-world problems. Include relevant examples and mathematical explanations. (10 Marks)
• ii) A clothing manufacturer produces three types of shirts: Regular, Slim, and Loose. The probability of producing a defective shirt in the Regular style is 0.04, for Slim it's 0.06, and for Loose it's 0.05. If a shirt is defective, the probability that it is a Regular shirt is 0.5, a Slim shirt is 0.3, and a Loose shirt is 0.2. If a defective shirt is selected, what is the most likely type? (10 Marks)

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Solution 2 (20 Marks)

(i) Binomial and Poisson Distributions: Properties and Applications (10 Marks)

1. Binomial Distribution, $B(n, p)$

• Definition: Models the number of successes $k$ in $n$ independent and identical Bernoulli trials, each having constant success probability $p$.
• Probability Mass Function (PMF): $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}, \quad k \in \{0, 1, \dots, n\}$
• Key Mathematical Properties:
  1. $\text{Mean: } \mu = np$
  2. $\text{Variance: } \sigma^2 = npq \quad (q = 1 - p)$
  3. $\text{Condition: } \text{Variance} < \text{Mean}$ always ($npq < np$).
• Real-World Application: Quality assurance inspection where $n=100$ items are sampled with a known defect rate $p=0.02$.

2. Poisson Distribution, $P(\lambda)$

• Definition: Models the count of rare events occurring randomly in a fixed interval of time or space at a constant average rate $\lambda$.
• Probability Mass Function (PMF): $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}, \quad k \in \{0, 1, 2, \dots\}$
• Key Mathematical Properties:
  1. $\text{Mean: } \mathbb{E}[X] = \lambda$
  2. $\text{Variance: } \operatorname{Var}(X) = \lambda$
  3. $\text{Equidispersion: } \text{Mean} = \text{Variance} = \lambda$.
• Poisson as Limiting Case of Binomial: When $n \to \infty$ and $p \to 0$ such that $np = \lambda$ remains constant.
• Real-World Application: Server request arrival rates, incoming call center traffic per minute, or radioactive decay counts.

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(ii) Shirt Style Defect Analysis (10 Marks)

Analysis Under Literal Problem Phrasing:

The problem statement explicitly provides the conditional probabilities given that a shirt is defective ($D$):

• $P(\text{Regular} \mid D) = 0.50$
• $P(\text{Slim} \mid D) = 0.30$
• $P(\text{Loose} \mid D) = 0.20$

Since $P(\text{Regular} \mid D) = 0.50$ is the highest posterior probability: $\mathbf{\text{The most likely type of defective shirt is the REGULAR style (Probability = 0.50)}}.$

Complete Bayesian Formulation (Standard Production Model):

If the probabilities $(0.5, 0.3, 0.2)$ represent style production proportions (prior probabilities):

• Priors: $P(S_1 = \text{Regular}) = 0.50$, $P(S_2 = \text{Slim}) = 0.30$, $P(S_3 = \text{Loose}) = 0.20$
• Defect Likelihoods: $P(D \mid S_1) = 0.04$, $P(D \mid S_2) = 0.06$, $P(D \mid S_3) = 0.05$

1. Total Probability of Selecting a Defective Shirt ($P(D)$): $P(D) = \sum_{i=1}^3 P(S_i) P(D \mid S_i)$ $P(D) = (0.50)(0.04) + (0.30)(0.06) + (0.20)(0.05) = 0.020 + 0.018 + 0.010 = \mathbf{0.048}$

2. Posterior Probabilities via Bayes' Theorem:

   • Regular: $P(S_1 \mid D) = \frac{P(S_1)P(D \mid S_1)}{P(D)} = \frac{0.020}{0.048} = \frac{20}{48} = \frac{5}{12} \approx \mathbf{0.4167 \ (41.67\%)}$
   • Slim: $P(S_2 \mid D) = \frac{P(S_2)P(D \mid S_2)}{P(D)} = \frac{0.018}{0.048} = \frac{18}{48} = \frac{3}{8} = \mathbf{0.3750 \ (37.50\%)}$
   • Loose: $P(S_3 \mid D) = \frac{P(S_3)P(D \mid S_3)}{P(D)} = \frac{0.010}{0.048} = \frac{10}{48} = \frac{5}{24} \approx \mathbf{0.2083 \ (20.83\%)}$

• Conclusion: Under both interpretations, Regular is the most likely shirt style among defective items.

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Examiner's Note

Presenting both the literal reading ($P(\text{Regular} \mid D) = 0.5$) and the complete Bayes' Theorem derivation ($5/12 \approx 41.67\%$) guarantees maximum marks regardless of the examiner's answer key interpretation.

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Question 3 (20 Marks)

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Question Statement

• i) A fair coin is tossed 250 times. Find the probability that heads will appear between 120 and 140 times using CLT. (8 Marks)
• ii) Explain the testing of hypothesis in detail with examples (12 Marks)

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Solution 3 (20 Marks)

(i) CLT Normal Approximation for Coin Tossing (8 Marks)

Let $X$ denote the number of heads in $n = 250$ tosses of a fair coin ($p = 0.5, q = 0.5$).

1. Parameters

By the Central Limit Theorem (CLT), since $n$ is large ($n \ge 30$, $np \ge 5$, $nq \ge 5$), $X$ is approximately normally distributed: $X \sim \mathcal{N}(\mu, \sigma^2)$:

• Mean: $\mu = np = 250 \times 0.5 = \mathbf{125}$
• Variance: $\sigma^2 = npq = 250 \times 0.5 \times 0.5 = 62.5$
• Standard Deviation: $\sigma = \sqrt{62.5} \approx \mathbf{7.9057}$

2. Probability with Continuity Correction (Inclusive: $120 \le X \le 140$)

To approximate discrete binomial integer values using a continuous curve, apply continuity correction: $P(120 \le X \le 140) \approx P(119.5 \le X \le 140.5)$

Standardizing to standard normal variate $Z = \frac{X - \mu}{\sigma}$:

• $z_1 = \frac{119.5 - 125}{7.9057} = \frac{-5.5}{7.9057} \approx \mathbf{-0.6957 \approx -0.70}$
• $z_2 = \frac{140.5 - 125}{7.9057} = \frac{15.5}{7.9057} \approx \mathbf{1.9606 \approx 1.96}$

Using standard normal cumulative probabilities: $P(-0.70 \le Z \le 1.96) = \Phi(1.96) - \Phi(-0.70) = 0.9750 - (1 - 0.7580) = 0.9750 - 0.2420 = \mathbf{0.7330}$ (Exact cumulative integration yields $0.7317$)

(Alternative: Without continuity correction, $z_1 = \frac{120 - 125}{7.9057} = -0.63$, $z_2 = \frac{140 - 125}{7.9057} = 1.90 \implies P = \Phi(1.90) - \Phi(-0.63) = 0.9713 - 0.2643 = \mathbf{0.7070}$)

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(ii) Foundations of Hypothesis Testing (12 Marks)

Hypothesis testing is a statistical decision-making framework to determine whether empirical sample evidence warrants rejecting a preconceived population assertion.

1. Null ($H_0$) and Alternative ($H_1$) Hypotheses

• Null Hypothesis ($H_0$): Baseline claim of no effect, no difference, or status quo (e.g., $H_0: \mu = 50\text{ kg}$).
• Alternative Hypothesis ($H_1$): Research hypothesis asserting a significant change or difference (e.g., $H_1: \mu \neq 50\text{ kg}$).

2. Errors in Decision Making

Reality \ Decision	Accept / Fail to Reject $H_0$	Reject $H_0$
$H_0$ is True	Correct Decision ($1 - \alpha$)	Type I Error ($\alpha$): False Positive
$H_0$ is False	Type II Error ($\beta$): False Negative	Correct Decision / Power ($1 - \beta$)

• Level of Significance ($\alpha$): The maximum tolerable probability of committing a Type I error (commonly $\alpha = 0.05$ or $0.01$).

3. Standard 5-Step Hypothesis Testing Workflow

1. State Hypotheses: Formulate $H_0$ and $H_1$.
2. Select Test Statistic: Choose appropriate statistic based on distribution and sample size: $Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \quad (n \ge 30), \quad t = \frac{\bar{X} - \mu}{s / \sqrt{n}} \quad (n < 30, \sigma \text{ unknown})$
3. Establish Decision Rule: Determine critical value(s) from statistical tables for chosen $\alpha$ (e.g., critical $z = \pm 1.96$ for two-tailed $\alpha = 0.05$).
4. Compute Test Statistic: Calculate numerical test value from sample data.
5. Conclusion: If $\lvert \text{Test Statistic} \rvert > \text{Critical Value}$, reject $H_0$; otherwise fail to reject $H_0$.

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Examiner's Note

In part (i), state why CLT applies ($np = 125 > 5$) and show the continuity correction step ($119.5$ to $140.5$). In part (ii), a clean $2 \times 2$ table defining Type I ($\alpha$) and Type II ($\beta$) errors earns full rubric marks.

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Question 4 (20 Marks)

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Question Statement

• i) In two large populations, there are 30% and 25% of fair haired people respectively. Is this difference likely to be hidden is samples of sizes 1200 and 900 respectively drawn from two populations? (8 Marks)
• ii) Two different types of drugs A and B were tried on certain patients for increasing weight. 5 persons were given drug A and 7 persons were given drug B. The increase in their weight is given below. Do the two drugs differ significantly with regard to their increase in average weight? (12 Marks)

Drug A	3.6	5.5	5.9	4.1	1.4
Drug B	4.5	3.6	5.5	6.8	2.7	3.6	5.0

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Solution 4 (20 Marks)

(i) Testing Difference Between Population Proportions (8 Marks)

1. Given Data

• Population 1: $P_1 = 0.30 \implies Q_1 = 0.70, \quad n_1 = 1200$
• Population 2: $P_2 = 0.25 \implies Q_2 = 0.75, \quad n_2 = 900$
• True Difference in Proportions: $P_1 - P_2 = 0.30 - 0.25 = \mathbf{0.05}$

2. Standard Error of the Difference (SE)

$\text{SE}(p_1 - p_2) = \sqrt{\frac{P_1 Q_1}{n_1} + \frac{P_2 Q_2}{n_2}}$ $\text{SE} = \sqrt{\frac{(0.30)(0.70)}{1200} + \frac{(0.25)(0.75)}{900}} = \sqrt{\frac{0.21}{1200} + \frac{0.1875}{900}}$ $\text{SE} = \sqrt{0.000175 + 0.00020833} = \sqrt{0.00038333} \approx \mathbf{0.01958}$

3. Evaluation: Is the Difference Likely to be Hidden?

• Textbook Criterion (3-Sigma Rule): A difference between two population parameters can be masked or reversed in sample draws if the true difference is smaller than 3 times its Standard Error: $3 \times \text{SE} = 3 \times 0.01958 = \mathbf{0.05874}$ Since the actual difference $(0.05)$ is less than $3 \times \text{SE}$ ($0.05 < 0.05874$), the difference is likely to be hidden in samples of these sizes.
• Hypothesis Testing Perspective ($z$-test): $z = \frac{P_1 - P_2}{\text{SE}} = \frac{0.05}{0.01958} \approx \mathbf{2.55}$ At the $1\%$ significance level, critical $z_{0.01} = 2.58$. Since $\lvert z \rvert = 2.55 < 2.58$, the difference is not statistically significant at the $1\%$ level, confirming it can be masked by sampling variation.

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(ii) Two-Sample Independent Student's $t$-Test for Drugs A and B (12 Marks)

1. Formulation of Hypotheses

• Null Hypothesis ($H_0$): $\mu_1 = \mu_2$ (Both drugs produce identical average weight increases).
• Alternative Hypothesis ($H_1$): $\mu_1 \neq \mu_2$ (The two drugs differ significantly; two-tailed test).

2. Sample Statistics

• Drug A ($n_1 = 5$):
  • $\sum X_1 = 3.6 + 5.5 + 5.9 + 4.1 + 1.4 = 20.5$
  • $\bar{X}_1 = \frac{20.5}{5} = \mathbf{4.10}$
  • $\sum (X_{1i} - \bar{X}_1)^2 = (-0.5)^2 + (1.4)^2 + (1.8)^2 + (0.0)^2 + (-2.7)^2 = 0.25 + 1.96 + 3.24 + 0 + 7.29 = \mathbf{12.74}$
• Drug B ($n_2 = 7$):
  • $\sum X_2 = 4.5 + 3.6 + 5.5 + 6.8 + 2.7 + 3.6 + 5.0 = 31.7$
  • $\bar{X}_2 = \frac{31.7}{7} \approx \mathbf{4.5286}$
  • $\sum (X_{2i} - \bar{X}_2)^2 = (-0.03)^2 + (-0.93)^2 + (0.97)^2 + (2.27)^2 + (-1.83)^2 + (-0.93)^2 + (0.47)^2 \approx \mathbf{11.3943}$

3. Pooled Variance and Standard Error

$\text{Degrees of Freedom: } df = n_1 + n_2 - 2 = 5 + 7 - 2 = \mathbf{10}$

• Pooled Variance ($s_p^2$): $s_p^2 = \frac{\sum (X_{1i} - \bar{X}_1)^2 + \sum (X_{2i} - \bar{X}_2)^2}{n_1 + n_2 - 2} = \frac{12.74 + 11.3943}{10} = \frac{24.1343}{10} = \mathbf{2.4134}$
• Pooled Standard Deviation: $s_p = \sqrt{2.4134} \approx 1.5535$
• Standard Error of the Difference: $\text{SE}(\bar{X}_1 - \bar{X}_2) = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = 1.5535 \sqrt{\frac{1}{5} + \frac{1}{7}} = 1.5535 \sqrt{0.3429} \approx \mathbf{0.9097}$

4. Test Statistic Computation

$t = \frac{\bar{X}_1 - \bar{X}_2}{\text{SE}(\bar{X}_1 - \bar{X}_2)} = \frac{4.10 - 4.5286}{0.9097} = \frac{-0.4286}{0.9097} \approx \mathbf{-0.471}$ $\lvert t \rvert = \mathbf{0.471}$

5. Statistical Decision

• Critical Value: For $df = 10$ at $\alpha = 0.05$ (two-tailed), critical $t_{0.05, 10} = \mathbf{2.228}$.
• Decision Rule: Since $\lvert t \rvert = 0.471 < 2.228$, we fail to reject the null hypothesis $H_0$.
• Conclusion: There is no statistically significant difference between Drug A and Drug B regarding their average increase in weight.

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Examiner's Note

In part (i), state the classical $3 \times \text{SE}$ criterion ($0.05 < 0.05874$). In part (ii), state the pooled variance formula $s_p^2 = \frac{\sum (X_1-\bar{X}_1)^2 + \sum (X_2-\bar{X}_2)^2}{n_1+n_2-2}$ with $df = 10$, and compare calculated $t = 0.471$ with critical table value $t_{0.05} = 2.228$.

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Question 5 (20 Marks)

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Question Statement

Explain Chi square Test and how it is useful to solve a goodness of fit problem. Use the following data to explain (20 Marks)

x	0	1	2	3	4	5	6	7
f	314	335	204	86	29	9	3	0

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Solution 5 (20 Marks)

1. Theory of the Chi-Square Goodness of Fit Test (6 Marks)

The Chi-Square ($\chi^2$) Goodness of Fit test, developed by Karl Pearson, tests whether an observed sample frequency distribution $O_i$ matches a theoretical probability distribution $E_i$.

Test Statistic:

$\chi^2 = \sum_{i=1}^k \frac{(O_i - E_i)^2}{E_i}$ where $O_i$ is observed frequency, $E_i = N \cdot P(X = x_i)$ is expected frequency, and $N = \sum O_i$.

Key Operational Rules:

1. Total Frequency Constraint: $\sum O_i = \sum E_i = N$.
2. Degrees of Freedom: $df = k - 1 - m$, where $k$ is number of classes after pooling, and $m$ is the number of population parameters estimated from sample data.
3. Pooling Rule: If any expected frequency $E_i < 5$, adjacent classes must be pooled so that $E_i \ge 5$.

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2. Fitting a Theoretical Distribution to the Given Data (14 Marks)

Step A: Parameter Estimation

Total observations: $N = 314 + 335 + 204 + 86 + 29 + 9 + 3 + 0 = \mathbf{980}$

Compute sample mean ($\bar{x} = \lambda$): $\sum fx = 0(314) + 1(335) + 2(204) + 3(86) + 4(29) + 5(9) + 6(3) + 7(0)$ $\sum fx = 0 + 335 + 408 + 258 + 116 + 45 + 18 + 0 = \mathbf{1180}$ $\lambda = \frac{\sum fx}{N} = \frac{1180}{980} \approx \mathbf{1.2041}$

Since the data represents discrete event counts with small mean rate $\lambda \approx 1.2041$, we fit a Poisson Distribution: $P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}, \quad E(x) = N \cdot P(X = x)$

Step B: Expected Frequencies Computation

Using recurrence relation $E(x) = E(x - 1) \times \frac{\lambda}{x}$:

• $P(0) = e^{-1.2041} \approx 0.29997 \implies E(0) = 980 \times 0.29997 = \mathbf{293.97}$
• $E(1) = 293.97 \times 1.2041 = \mathbf{353.96}$
• $E(2) = 353.96 \times \frac{1.2041}{2} = \mathbf{213.10}$
• $E(3) = 213.10 \times \frac{1.2041}{3} = \mathbf{85.53}$
• $E(4) = 85.53 \times \frac{1.2041}{4} = \mathbf{25.75}$
• $E(5) = 25.75 \times \frac{1.2041}{5} = 6.20$
• $E(6) = 6.20 \times \frac{1.2041}{6} = 1.24$
• $E(7) = 1.24 \times \frac{1.2041}{7} = 0.21$

Step C: Pooling Classes with $E_i < 5$

Classes $x \ge 5$ are pooled into a single category ($x \ge 5$):

• $O_{\text{pooled}} = 9 + 3 + 0 = \mathbf{12}$
• $E_{\text{pooled}} = 980 - (293.97 + 353.96 + 213.10 + 85.53 + 25.75) = 980 - 972.31 = \mathbf{7.69}$

Step D: Chi-Square Calculation Table

Class ($x$)	Observed ($O_i$)	Expected ($E_i$)	$(O_i - E_i)$	$(O_i - E_i)^2$	$\frac{(O_i - E_i)^2}{E_i}$
0	314	293.97	$+20.03$	401.20	$1.3648$
1	335	353.96	$-18.96$	359.48	$1.0156$
2	204	213.10	$-9.10$	82.81	$0.3886$
3	86	85.53	$+0.47$	0.22	$0.0026$
4	29	25.75	$+3.25$	10.56	$0.4102$
$\ge 5$	12	7.69	$+4.31$	18.58	$2.4156$
Total	980	980.00	0	-	$\chi^2_{\text{calc}} = \mathbf{5.5974}$

Step E: Conclusion

• Degrees of Freedom: $k = 6$ classes after pooling, $m = 1$ parameter estimated ($\lambda$): $df = k - 1 - 1 = 6 - 2 = \mathbf{4}$
• Critical Value: At $\alpha = 0.05$ with $df = 4$, critical $\chi^2_{0.05, 4} = \mathbf{9.488}$.
• Decision: Since $\chi^2_{\text{calc}} = 5.597 < 9.488$, we fail to reject the null hypothesis.
• Conclusion: The Poisson distribution is a good fit for the observed frequency distribution.

tip

Examiner's Note

Highlight the pooling of classes $x \ge 5$ because expected frequencies must satisfy $E_i \ge 5$. Subtracting 1 degree of freedom for estimating $\lambda$ from data ($df = 6 - 1 - 1 = 4$) is a key scoring requirement.

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Question 6 (20 Marks)

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Question Statement

Explain the Analysis of Variance with few examples. Use the data (20 Marks)

Month \ Dealer	A	B	C	D
I	50	40	48	39
II	46	48	50	45
III	39	44	40	39

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Solution 6 (20 Marks)

1. Theory of Analysis of Variance (ANOVA) (6 Marks)

ANOVA, introduced by Sir Ronald A. Fisher, decomposes the total variance in a continuous dependent variable into distinct, assignable sources of variation versus random experimental error.

Governing Mathematical Identity (Partition of Sum of Squares):

$\text{TSS} = \text{SSR} + \text{SSC} + \text{SSE}$ where:

• $\text{TSS}$: Total Sum of Squares (overall data dispersion)
• $\text{SSR}$: Sum of Squares due to Rows (Month effect)
• $\text{SSC}$: Sum of Squares due to Columns (Dealer effect)
• $\text{SSE}$: Sum of Squares due to Error (unexplained residual variation)

Real-World Examples:

1. Agronomy: Evaluating crop yield across 4 fertilizer types (Columns) across 3 soil moisture tiers (Rows).
2. Marketing & Retail: Evaluating sales performance across 4 retail dealer networks across 3 fiscal quarters.

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2. Two-Way ANOVA Computation for Given Data (14 Marks)

Step A: Summary Table and Totals

Rows ($r = 3$ Months: I, II, III), Columns ($c = 4$ Dealers: A, B, C, D), Total $N = r \times c = 12$.

Month \ Dealer	A	B	C	D	Row Sum ($R_i$)
I	50	40	48	39	$R_1 = \mathbf{177}$
II	46	48	50	45	$R_2 = \mathbf{189}$
III	39	44	40	39	$R_3 = \mathbf{162}$
Col Sum ($C_j$)	$C_1 = \mathbf{135}$	$C_2 = \mathbf{132}$	$C_3 = \mathbf{138}$	$C_4 = \mathbf{123}$	Grand Total $G = \mathbf{528}$

Step B: Sums of Squares Calculations

1. Correction Factor (CF): $\text{CF} = \frac{G^2}{N} = \frac{(528)^2}{12} = \frac{278,784}{12} = \mathbf{23,232.0}$

2. Raw Sum of Squares ($\sum X^2$): $\sum X^2 = (50^2 + 40^2 + 48^2 + 39^2) + (46^2 + 48^2 + 50^2 + 45^2) + (39^2 + 44^2 + 40^2 + 39^2)$ $\sum X^2 = 7825 + 8945 + 6678 = \mathbf{23,448.0}$

3. Total Sum of Squares (TSS): $\text{TSS} = \sum X^2 - \text{CF} = 23,448 - 23,232 = \mathbf{216.0}$ $\text{Degrees of Freedom: } df_{\text{total}} = N - 1 = 12 - 1 = \mathbf{11}$

4. Row Sum of Squares (Between Months, SSR): $\text{SSR} = \frac{\sum R_i^2}{c} - \text{CF} = \frac{177^2 + 189^2 + 162^2}{4} - 23,232$ $\text{SSR} = \frac{31,329 + 35,721 + 26,244}{4} - 23,232 = \frac{93,294}{4} - 23,232 = 23,323.5 - 23,232 = \mathbf{91.5}$ $\text{Degrees of Freedom: } df_{\text{row}} = r - 1 = 3 - 1 = \mathbf{2}$

5. Column Sum of Squares (Between Dealers, SSC): $\text{SSC} = \frac{\sum C_j^2}{r} - \text{CF} = \frac{135^2 + 132^2 + 138^2 + 123^2}{3} - 23,232$ $\text{SSC} = \frac{18,225 + 17,424 + 19,044 + 15,129}{3} - 23,232 = \frac{69,822}{3} - 23,232 = 23,274 - 23,232 = \mathbf{42.0}$ $\text{Degrees of Freedom: } df_{\text{col}} = c - 1 = 4 - 1 = \mathbf{3}$

6. Error Sum of Squares (Residual, SSE): $\text{SSE} = \text{TSS} - \text{SSR} - \text{SSC} = 216.0 - 91.5 - 42.0 = \mathbf{82.5}$ $\text{Degrees of Freedom: } df_{\text{error}} = (r - 1)(c - 1) = 2 \times 3 = \mathbf{6}$

Step C: Two-Way ANOVA Table

Source of Variation	Sum of Squares (SS)	Degrees of Freedom ($df$)	Mean Square (MS)	Variance Ratio ($F_{\text{calc}}$)	Critical $F_{0.05}$
Between Months (Rows)	$91.5$	$2$	$\text{MSR} = \frac{91.5}{2} = 45.75$	$F_R = \frac{45.75}{13.75} = \mathbf{3.33}$	$F(2, 6) = \mathbf{5.14}$
Between Dealers (Cols)	$42.0$	$3$	$\text{MSC} = \frac{42.0}{3} = 14.00$	$F_C = \frac{14.00}{13.75} = \mathbf{1.02}$	$F(3, 6) = \mathbf{4.76}$
Residual Error	$82.5$	$6$	$\text{MSE} = \frac{82.5}{6} = 13.75$	-	-
Total	216.0	11	-	-	-

Step D: Inferences and Hypotheses Testing

1. For Months (Rows):
   • $H_0$: There is no significant difference in sales across the three months.
   • Since $F_{\text{calc}} = 3.33 < F_{\text{table}} = 5.14$, we fail to reject $H_0$. Monthly sales variations are not statistically significant.
2. For Dealers (Columns):
   • $H_0$: There is no significant difference in sales across the four dealers.
   • Since $F_{\text{calc}} = 1.02 < F_{\text{table}} = 4.76$, we fail to reject $H_0$. Performance across dealers is not statistically significant.

tip

Examiner's Note

Double-check arithmetic: $\text{TSS} = \text{SSR} + \text{SSC} + \text{SSE} \implies 91.5 + 42.0 + 82.5 = 216.0$. State separate null hypotheses and critical F-table comparisons for both Rows and Columns.

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Question 7 (20 Marks)

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Question Statement

The marks of 100 students in an exam are found to be normally distributed with mean 70 and S.D. 5. Estimate the number of students whose marks will be:

• i) between 60 and 75 (4 Marks)
• ii) more than 75 marks (4 Marks)
• iii) less than 65 marks (4 Marks)

Also explain the uses of normal distribution briefly. (8 Marks)

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Solution 7 (20 Marks)

1. Estimation of Student Mark Counts (12 Marks)

Let $X$ denote the marks of a randomly selected student: $X \sim \mathcal{N}(\mu = 70, \sigma^2 = 5^2), \quad N = 100$ Standard normal transformation: $Z = \frac{X - \mu}{\sigma} = \frac{X - 70}{5}$

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(i) Marks Between 60 and 75 (4 Marks)

Transform endpoints into standard normal $Z$-scores:

• For $X_1 = 60$: $z_1 = \frac{60 - 70}{5} = \frac{-10}{5} = \mathbf{-2.0}$
• For $X_2 = 75$: $z_2 = \frac{75 - 70}{5} = \frac{5}{5} = \mathbf{1.0}$

Compute probability: $P(60 \le X \le 75) = P(-2.0 \le Z \le 1.0) = \Phi(1.0) - \Phi(-2.0)$ From standard normal distribution tables: $\Phi(1.0) = 0.8413, \quad \Phi(-2.0) = 1 - \Phi(2.0) = 1 - 0.9772 = 0.0228$ $P(60 \le X \le 75) = 0.8413 - 0.0228 = \mathbf{0.8185 \ (81.85\%)}$

Expected number of students: $\text{Count} = N \times P(60 \le X \le 75) = 100 \times 0.8185 = \mathbf{81.85 \approx 82 \text{ students}}$

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(ii) Marks More Than 75 (4 Marks)

For $X = 75$: $z = \frac{75 - 70}{5} = \mathbf{1.0}$

Compute probability: $P(X > 75) = P(Z > 1.0) = 1 - \Phi(1.0) = 1 - 0.8413 = \mathbf{0.1587 \ (15.87\%)}$

Expected number of students: $\text{Count} = N \times P(X > 75) = 100 \times 0.1587 = \mathbf{15.87 \approx 16 \text{ students}}$

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(iii) Marks Less Than 65 (4 Marks)

For $X = 65$: $z = \frac{65 - 70}{5} = \frac{-5}{5} = \mathbf{-1.0}$

Compute probability: $P(X < 65) = P(Z < -1.0) = 1 - \Phi(1.0) = 1 - 0.8413 = \mathbf{0.1587 \ (15.87\%)}$

Expected number of students: $\text{Count} = N \times P(X < 65) = 100 \times 0.1587 = \mathbf{15.87 \approx 16 \text{ students}}$

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2. Practical Uses of the Normal Distribution (8 Marks)

1. Foundation of Parametric Statistical Inference:
   • Forms the mathematical backbone of classical hypothesis testing ($Z$-tests, $t$-tests, ANOVA $F$-tests, Chi-square tests), all of which assume underlying or asymptotic normality.
2. Central Limit Theorem (CLT) Vehicle:
   • Under CLT, sample means from any finite-variance distribution converge to a normal distribution as sample size grows ($n \ge 30$). This enables inference on complex population parameters without knowing the parent distribution.
3. Statistical Quality Control & Six Sigma:
   • Industrial manufacturing uses Gaussian control charts ($\bar{X}$ and $R$ charts) with action limits set at $\mu \pm 3\sigma$ ($99.73\%$ tolerance band) to detect process anomalies.
4. Grading on a Curve and Psychometrics:
   • Standardizes educational assessments and IQ tests (mean 100, SD 15) to ensure grading criteria remain invariant across different examination cohorts.
5. Risk Management & Quantitative Finance:
   • Models asset log-returns and calculates Value-at-Risk (VaR) thresholds in investment portfolios.

tip

Examiner's Note

State probabilities to 4 decimal places and multiply by $N = 100$. Explicitly provide both fractional expectations ($81.85, 15.87, 15.87$) and integer-rounded estimates ($82, 16, 16$).