🧮 How to Use This Section

This section contains solved numerical problems from the major topics of B.Sc. 2nd Year Applied Statistics. Each problem is presented step-by-step so that students can understand both the method and the calculation.

Standard pattern: Question → Given → Formula → Substitution → Calculation → Final Answer → Interpretation

Unit 1: Methods of Data Summarization

🟢 Basic

Numerical 1: Arithmetic Mean

Question: Calculate the arithmetic mean of the observations: 5, 8, 10, 12 and 15.

Given

x = 5, 8, 10, 12, 15

Formula
x̄ = Σx / n
Substitution

Σx = 5 + 8 + 10 + 12 + 15 = 50

n = 5

Calculation
x̄ = 50 / 5 = 10
Final Answer: Arithmetic mean = 10
Interpretation: The average value of the observations is 10.
🟢 Basic

Numerical 2: Median

Question: Find the median of: 4, 7, 2, 9, 6, 5, 8.

Given

Data = 4, 7, 2, 9, 6, 5, 8

Step 1: Arrange the data

2, 4, 5, 6, 7, 8, 9

Formula
Position of Median = (n + 1) / 2
Calculation
= (7 + 1) / 2 = 4th observation
Final Answer: Median = 6
🟡 Moderate

Numerical 3: Standard Deviation

Question: Calculate the standard deviation of: 2, 4, 6, 8 and 10.

Given

x = 2, 4, 6, 8, 10

Step 1: Calculate Mean
x̄ = (2 + 4 + 6 + 8 + 10) / 5 = 6
x x − x̄ (x − x̄)²
2 -4 16
4 -2 4
6 0 0
8 2 4
10 4 16
Total 40
Formula
σ = √[Σ(x − x̄)² / n]
Calculation
σ = √(40 / 5)
σ = √8
σ = 2.83
Final Answer: Standard deviation ≈ 2.83
Interpretation: The observations have a standard deviation of approximately 2.83 units from their mean.
🟡 Moderate

Numerical 4: Coefficient of Variation

Question: The mean of a dataset is 50 and its standard deviation is 5. Calculate the coefficient of variation.

Given

Mean = 50
SD = 5

Formula
CV = (SD / Mean) × 100
Calculation
CV = (5 / 50) × 100
CV = 10%
Final Answer: Coefficient of variation = 10%
Interpretation: The relative variability of the dataset is 10%.
🔴 Exam Level

Numerical 5: Mean from Frequency Distribution

Question: Calculate the mean from the following frequency distribution.

x f fx
10 2 20
20 3 60
30 4 120
40 1 40
Total 10 240
Formula
x̄ = Σfx / Σf
Calculation
x̄ = 240 / 10
x̄ = 24
Final Answer: Mean = 24

Unit 2: Correlation

🟡 Moderate

Numerical 1: Pearson's Correlation

Question: Calculate Karl Pearson's correlation coefficient for:

X 1 2 3 4 5
Y 2 4 5 4 6
Formula
r = [nΣxy − ΣxΣy] / √{[nΣx² − (Σx)²][nΣy² − (Σy)²]}
Required values

n = 5
Σx = 15
Σy = 21
Σx² = 55
Σy² = 97
Σxy = 73

Substitution
r = [5(73) − (15)(21)] / √{[5(55) − 15²][5(97) − 21²]}
r = 50 / √(50 × 44)

r ≈ 0.95
Final Answer: Pearson's correlation coefficient ≈ +0.95
Interpretation: There is a strong positive linear correlation between X and Y.
🟢 Basic

Numerical 2: Spearman's Rank Correlation

Question: Calculate Spearman's rank correlation coefficient when the ranks are:

X 1 2 3 4 5
Y 2 1 3 5 4
Formula
ρ = 1 − [6Σd² / n(n² − 1)]

Differences in ranks:

d -1 1 0 -1 1
1 1 0 1 1

Σd² = 4

ρ = 1 − [6(4) / 5(25 − 1)]

ρ = 1 − 24/120

ρ = 0.80
Final Answer: Spearman's rank correlation = 0.80
Interpretation: The two rankings show a strong positive association.
🔴 Exam Level

Numerical 3: Interpretation of Correlation

Question: A study reports a correlation coefficient of r = −0.82 between two biological variables. Interpret the result.

Interpretation

The negative sign indicates an inverse relationship between the two variables. The magnitude, 0.82, indicates a strong linear association.

Final Answer: There is a strong negative linear correlation between the two variables.
Correlation indicates association, not necessarily causation.

Unit 3: Methods of Data Modeling

🟡 Moderate

Numerical 1: Simple Linear Regression

Question: For the regression equation

Y = 5 + 2X

predict Y when X = 10.

Given

a = 5
b = 2
X = 10

Substitution
Y = 5 + 2(10)
Calculation
Y = 5 + 20
Y = 25
Final Answer: Predicted Y = 25
Interpretation: When X is 10, the estimated value of Y is 25.
🟢 Basic

Numerical 2: Residual

Question: If the observed value is 28 and the predicted value is 25, calculate the residual.

Formula
Residual = Observed − Predicted
Residual = 28 − 25 = 3
Final Answer: Residual = 3
🟢 Basic

Numerical 3: Coefficient of Determination

Question: If the correlation coefficient between X and Y is 0.8, calculate R² and interpret it.

Formula
R² = r²
Calculation
R² = (0.8)² = 0.64
Final Answer: R² = 0.64 or 64%
Interpretation: About 64% of the variation in the dependent variable is explained by the linear relationship with the predictor in this simple regression setting.

Unit 4: Analysis of Categorical Data

🟡 Moderate

Numerical 1: Expected Frequency

Question: In a contingency table, the row total is 40, the column total is 30, and the grand total is 100. Calculate the expected frequency.

Formula
E = (Row Total × Column Total) / Grand Total
Substitution
E = (40 × 30) / 100
E = 12
Final Answer: Expected frequency = 12
🟡 Moderate

Numerical 2: Chi-square Statistic

Question: If observed frequency is 18 and expected frequency is 15, calculate the contribution of the cell to χ².

Formula
χ² contribution = (O − E)² / E
Substitution
= (18 − 15)² / 15
= 9 / 15
= 0.60
Final Answer: Contribution = 0.60
🟢 Basic

Numerical 3: Yule's Coefficient

Question: For a 2 × 2 table, let a = 20, b = 10, c = 5 and d = 15. Calculate Yule's coefficient of association.

Formula
Q = (ad − bc) / (ad + bc)
Substitution
Q = [(20)(15) − (10)(5)] / [(20)(15) + (10)(5)]
Q = (300 − 50)/(300 + 50)
Q = 250/350
Q ≈ 0.714
Final Answer: Q ≈ +0.71
Interpretation: The variables show a positive association.

Unit 5: Introduction to Probability

🟢 Basic

Numerical 1: Simple Probability

Question: A box contains 10 bacterial culture tubes, of which 4 are positive. If one tube is selected randomly, find the probability that it is positive.

Formula
P(A) = Favorable outcomes / Total outcomes
P(A) = 4 / 10 = 0.4
Final Answer: P(A) = 0.4 or 40%
🟡 Moderate

Numerical 2: Addition Theorem

Question: If P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2, calculate P(A ∪ B).

Formula
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Calculation
= 0.5 + 0.4 − 0.2
= 0.7
Final Answer: P(A ∪ B) = 0.70
🟡 Moderate

Numerical 3: Conditional Probability

Question: If P(A ∩ B) = 0.20 and P(B) = 0.50, find P(A|B).

Formula
P(A|B) = P(A ∩ B) / P(B)
P(A|B) = 0.20 / 0.50
= 0.40
Final Answer: P(A|B) = 0.40
🔴 Exam Level

Numerical 4: Bayes' Theorem

Question: A disease occurs in 10% of a population. A test is positive in 90% of diseased individuals and in 5% of non-diseased individuals. Find the probability that a person actually has the disease given that the test is positive.

Given

P(D) = 0.10
P(+|D) = 0.90
P(+|D̄) = 0.05
P(D̄) = 0.90

Formula
P(D|+) = [P(+|D)P(D)] / [P(+|D)P(D) + P(+|D̄)P(D̄)]
Substitution
P(D|+) = (0.90 × 0.10) / [(0.90 × 0.10) + (0.05 × 0.90)]

= 0.09 / 0.135

= 0.667
Final Answer: P(D|+) ≈ 0.667 or 66.7%
Interpretation: Under the stated assumptions, among people with a positive test, the estimated probability of actually having the disease is about 66.7%.

Unit 6: Probability Distributions

🟡 Moderate

Numerical 1: Binomial Distribution

Question: If the probability of success is 0.4, calculate the probability of exactly 2 successes in 5 independent trials.

Given

n = 5
x = 2
p = 0.4
q = 0.6

Formula
P(X=x) = ⁿCₓ pˣqⁿ⁻ˣ
Substitution
P(X=2) = ⁵C₂(0.4)²(0.6)³
= 10 × 0.16 × 0.216
= 0.3456
Final Answer: P(X=2) = 0.3456
🟢 Basic

Numerical 2: Binomial Mean and Variance

Question: For a binomial distribution with n = 20 and p = 0.3, calculate the mean and variance.

Mean
Mean = np
= 20 × 0.3
= 6
Variance
Variance = npq
= 20 × 0.3 × 0.7
= 4.2
Final Answer: Mean = 6
Variance = 4.2
🟡 Moderate

Numerical 3: Poisson Distribution

Question: If the average number of bacterial colonies observed in a given sample is 3, calculate the probability of observing exactly 2 colonies under a Poisson model.

Given

λ = 3
x = 2

Formula
P(X=x) = e⁻λ λˣ / x!
Substitution
P(X=2) = e⁻³(3²)/2!
≈ 0.224
Final Answer: P(X=2) ≈ 0.224
🟡 Moderate

Numerical 4: Normal Distribution

Question: A variable has mean 50 and standard deviation 10. Find the Z-score corresponding to X = 70.

Formula
Z = (X − μ) / σ
Substitution
Z = (70 − 50) / 10
Z = 2
Final Answer: Z = 2.0
Interpretation: The value 70 is two standard deviations above the mean.

Unit 7: Estimation

🟡 Moderate

Numerical 1: Confidence Interval for Mean

Question: A sample has mean 50, population standard deviation 10 and sample size 100. Calculate the 95% confidence interval for the population mean.

Given

x̄ = 50
σ = 10
n = 100
Z = 1.96

Formula
CI = x̄ ± Z(σ/√n)
Substitution
CI = 50 ± 1.96(10/√100)
= 50 ± 1.96
Final Answer: 95% CI = 48.04 to 51.96
Interpretation: The calculated 95% confidence interval for the population mean is 48.04 to 51.96.
🟢 Basic

Numerical 2: Confidence Interval for Proportion

Question: In a sample of 100 individuals, 60 are positive for a particular test. Calculate the approximate 95% confidence interval for the population proportion.

Given

p = 60/100 = 0.60
q = 0.40
n = 100
Z = 1.96

Formula
CI = p ± Z√(pq/n)
Calculation
SE = √[(0.60 × 0.40)/100]
SE ≈ 0.049
CI = 0.60 ± 1.96(0.049)
CI ≈ 0.60 ± 0.096
Final Answer: 95% CI ≈ 0.504 to 0.696

or approximately 50.4% to 69.6%.

Unit 8: Hypothesis Testing

🟡 Moderate

Numerical 1: One-Sample Z Test

Question: A population has hypothesized mean 50 and known standard deviation 10. A sample of 100 has mean 52. Test the null hypothesis at the 5% significance level.

Given

μ = 50
x̄ = 52
σ = 10
n = 100
α = 0.05

Hypotheses

H₀: μ = 50
H₁: μ ≠ 50

Formula
Z = (x̄ − μ)/(σ/√n)
Calculation
Z = (52 − 50)/(10/√100)
Z = 2

At 5% significance level for a two-tailed test, the critical value is approximately ±1.96.

Decision: Since |2| > 1.96, reject H₀.
Interpretation: There is statistically significant evidence at the 5% level that the population mean differs from 50.
🟡 Moderate

Numerical 2: One-Sample t Test

Question: A sample of 10 observations has mean 24 and standard deviation 4. Test whether the population mean differs from 20 at the 5% level.

Given

x̄ = 24
μ = 20
s = 4
n = 10

Hypotheses

H₀: μ = 20
H₁: μ ≠ 20

Formula
t = (x̄ − μ)/(s/√n)
Calculation
t = (24 − 20)/(4/√10)
t ≈ 3.16

Degrees of freedom:

df = n − 1 = 9
Final Answer: Calculated t ≈ 3.16.
The calculated statistic should be compared with the appropriate two-tailed t critical value for 9 degrees of freedom at α = 0.05.
🟡 Moderate

Numerical 3: F Test

Question: Two samples have variances 25 and 10. Calculate the F statistic by placing the larger variance in the numerator.

Formula
F = Larger variance / Smaller variance
F = 25 / 10
F = 2.5
Final Answer: F = 2.5
The calculated F value must be compared with the appropriate critical F value using the relevant numerator and denominator degrees of freedom.
🔴 Exam Level

Numerical 4: One-Way ANOVA

Question: Three groups have the following observations:

Group A Group B Group C
8 12 15
9 11 14
10 13 16

The purpose of ANOVA is to determine whether the group means differ significantly.

Step 1: Group Means

Group A mean = 9
Group B mean = 12
Group C mean = 15

Overall Mean
Overall Mean = (9 + 12 + 15)/3 = 12

The complete ANOVA calculation requires the between-group and within-group sums of squares.

Basic ANOVA Formula
F = MS Between / MS Within
Final Answer: ANOVA is used to test whether the observed differences among the three group means are statistically significant.
Interpretation: The calculated F statistic must be compared with the appropriate critical F value, or interpreted using the p-value.

Unit 9: Non-Parametric Tests

🟢 Basic

Numerical 1: Sign Test

Question: In a paired study, 8 observations show a positive difference and 2 show a negative difference. Ignoring ties, determine the number of positive and negative signs.

Given

Positive differences = 8
Negative differences = 2
Total = 10

Final Answer: Number of positive signs = 8
Number of negative signs = 2
These counts are then used in the appropriate binomial-based sign-test procedure.
🟡 Moderate

Numerical 2: Wilcoxon Signed-Rank Test

Question: Consider paired differences:

+2, −1, +4, +3, −2

Step 1: Ignore zero differences

There are no zero differences.

Step 2: Rank absolute differences
Difference |Difference| Rank Sign
+2 2 2.5 +
−1 1 1
+4 4 5 +
+3 3 4 +
−2 2 2.5
Step 3: Sum the ranks

Positive rank sum:

W⁺ = 2.5 + 5 + 4 = 11.5

Negative rank sum:

W⁻ = 1 + 2.5 = 3.5
Final Answer: W⁺ = 11.5
W⁻ = 3.5
The smaller signed-rank sum is used as the test statistic, subject to the appropriate Wilcoxon critical-value procedure.
🟡 Moderate

Numerical 3: Mann–Whitney U Test

Question: Two independent groups have the following observations:

Group A: 4, 7, 8
Group B: 2, 5, 6

Step 1: Combine and rank
Value Group Rank
2 B 1
4 A 2
5 B 3
6 B 4
7 A 5
8 A 6
Rank sums

R₁ for Group A = 2 + 5 + 6 = 13

R₂ for Group B = 1 + 3 + 4 = 8

Formula
U₁ = n₁n₂ + [n₁(n₁+1)/2] − R₁
U₁ = (3)(3) + [3(4)/2] − 13
= 9 + 6 − 13
= 2
Final Answer: U₁ = 2
The calculated U statistic is compared with the appropriate critical value or evaluated using the p-value.
🔴 Exam Level

Numerical 4: Kruskal–Wallis Test

Question: Three independent groups contain the following observations:

Group A Group B Group C
2 5 8
3 6 9
4 7 10
Step 1: Rank all observations

The combined observations are: 2, 3, 4, 5, 6, 7, 8, 9, 10.

Their ranks are therefore 1 through 9.

Rank sums

Group A:

R₁ = 1 + 2 + 3 = 6

Group B:

R₂ = 4 + 5 + 6 = 15

Group C:

R₃ = 7 + 8 + 9 = 24
Formula
H = [12 / N(N+1)] Σ(Rᵢ²/nᵢ) − 3(N+1)

Here N = 9 and each group contains 3 observations.

H = [12/(9×10)] [ 6²/3 + 15²/3 + 24²/3 ] − 30

H = 0.1333(12 + 75 + 192) − 30

H = 0.1333(279) − 30

H ≈ 7.2
Final Answer: Kruskal–Wallis statistic H ≈ 7.2
The calculated H statistic should be compared with the appropriate chi-square critical value with k − 1 degrees of freedom, or interpreted using the p-value.

🎯 Final Revision Checklist

Unit Important Numerical Areas
Unit 1 Mean, Median, Mode, Variance, SD, CV, Quartiles
Unit 2 Pearson, Spearman, Tied Ranks, Kendall's Tau
Unit 3 Regression, Prediction, Residual, R²
Unit 4 Expected Frequency, Chi-square, Yule's Q, Contingency
Unit 5 Addition, Multiplication, Conditional Probability, Bayes
Unit 6 Binomial, Poisson, Normal Distribution
Unit 7 Confidence Interval, Proportion, Sample Size
Unit 8 Z, t, F, ANOVA, Correlation and Regression Tests
Unit 9 Sign, Wilcoxon, Mann–Whitney U, Kruskal–Wallis