Unit 1: Solved Numerical Problems

This section contains 20+ fully solved problems covering all topics from Unit 1. Each solution includes detailed step-by-step explanations.

Section A: Measures of Central Tendency

Problem 1: Arithmetic Mean (Ungrouped Data)

Question: The monthly salaries (in thousands) of 10 government employees are: 25, 30, 28, 35, 40, 32, 38, 45, 29, 33

Calculate the arithmetic mean salary.

Click to reveal solution **Step 1:** Identify the data - Values: 25, 30, 28, 35, 40, 32, 38, 45, 29, 33 - Number of observations (n) = 10 **Step 2:** Apply the formula $$\bar{x} = \frac{\sum x}{n}$$ **Step 3:** Calculate sum of all values $$\sum x = 25 + 30 + 28 + 35 + 40 + 32 + 38 + 45 + 29 + 33 = 335$$ **Step 4:** Calculate mean $$\bar{x} = \frac{335}{10} = 33.5$$ **Answer:** The arithmetic mean salary is **Rs. 33,500** (or 33.5 thousand).

Problem 2: Arithmetic Mean (Grouped Data - Direct Method)

Question: The following table shows the distribution of marks obtained by 50 students:

Marks 10-20 20-30 30-40 40-50 50-60
No. of Students 5 10 18 12 5

Calculate the mean marks using the direct method.

Click to reveal solution **Step 1:** Create a table with mid-points | Marks | Mid-point (x) | Frequency (f) | fx | | --------- | ------------- | ------------- | -------- | | 10-20 | 15 | 5 | 75 | | 20-30 | 25 | 10 | 250 | | 30-40 | 35 | 18 | 630 | | 40-50 | 45 | 12 | 540 | | 50-60 | 55 | 5 | 275 | | **Total** | | **50** | **1770** | **Step 2:** Calculate mid-points - Mid-point = (Lower limit + Upper limit) / 2 - For 10-20: (10 + 20) / 2 = 15 - For 20-30: (20 + 30) / 2 = 25 - And so on... **Step 3:** Calculate fx for each class - For 10-20: 15 × 5 = 75 - For 20-30: 25 × 10 = 250 - For 30-40: 35 × 18 = 630 - For 40-50: 45 × 12 = 540 - For 50-60: 55 × 5 = 275 **Step 4:** Apply the formula $$\bar{x} = \frac{\sum fx}{\sum f} = \frac{1770}{50} = 35.4$$ **Answer:** The mean marks is **35.4**

Problem 3: Arithmetic Mean (Assumed Mean Method)

Question: Using the data from Problem 2, calculate mean using the assumed mean method with A = 35.

Click to reveal solution **Step 1:** Create deviation table (d = x - A, where A = 35) | Marks | x | f | d = x - 35 | fd | | --------- | --- | ------ | ---------- | ------ | | 10-20 | 15 | 5 | -20 | -100 | | 20-30 | 25 | 10 | -10 | -100 | | 30-40 | 35 | 18 | 0 | 0 | | 40-50 | 45 | 12 | 10 | 120 | | 50-60 | 55 | 5 | 20 | 100 | | **Total** | | **50** | | **20** | **Step 2:** Apply the assumed mean formula $$\bar{x} = A + \frac{\sum fd}{\sum f}$$ **Step 3:** Substitute values $$\bar{x} = 35 + \frac{20}{50} = 35 + 0.4 = 35.4$$ **Answer:** The mean marks is **35.4** (same as direct method)

Problem 4: Arithmetic Mean (Step Deviation Method)

Question: Using the same data, calculate mean using step deviation method with A = 35 and h = 10.

Click to reveal solution **Step 1:** Create step deviation table (u = (x - A)/h) | Marks | x | f | u = (x-35)/10 | fu | | --------- | --- | ------ | ------------- | ----- | | 10-20 | 15 | 5 | -2 | -10 | | 20-30 | 25 | 10 | -1 | -10 | | 30-40 | 35 | 18 | 0 | 0 | | 40-50 | 45 | 12 | 1 | 12 | | 50-60 | 55 | 5 | 2 | 10 | | **Total** | | **50** | | **2** | **Step 2:** Apply step deviation formula $$\bar{x} = A + h \times \frac{\sum fu}{\sum f}$$ **Step 3:** Substitute values $$\bar{x} = 35 + 10 \times \frac{2}{50} = 35 + 10 \times 0.04 = 35 + 0.4 = 35.4$$ **Answer:** The mean marks is **35.4**

Problem 5: Weighted Mean

Question: A student scored the following marks in different subjects:

Subject Marks Credit Hours
Statistics 75 3
Economics 68 4
Public Admin 82 3
Research Methods 70 2

Calculate the weighted mean (GPA calculation).

Click to reveal solution **Step 1:** Set up the calculation table | Subject | Marks (x) | Weight (w) | wx | | ---------------- | --------- | ---------- | ------- | | Statistics | 75 | 3 | 225 | | Economics | 68 | 4 | 272 | | Public Admin | 82 | 3 | 246 | | Research Methods | 70 | 2 | 140 | | **Total** | | **12** | **883** | **Step 2:** Apply weighted mean formula $$\bar{x}_w = \frac{\sum wx}{\sum w}$$ **Step 3:** Calculate $$\bar{x}_w = \frac{883}{12} = 73.58$$ **Answer:** The weighted mean score is **73.58 marks**

Problem 6: Combined Mean

Question: Two sections of MPA students have the following results:

  • Section A: 40 students, mean marks = 65
  • Section B: 35 students, mean marks = 72

Find the combined mean marks of both sections.

Click to reveal solution **Step 1:** Identify given values - $n_1 = 40$, $\bar{x}_1 = 65$ - $n_2 = 35$, $\bar{x}_2 = 72$ **Step 2:** Apply combined mean formula $$\bar{x}_{combined} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$$ **Step 3:** Calculate $$\bar{x}_{combined} = \frac{40 \times 65 + 35 \times 72}{40 + 35}$$ $$= \frac{2600 + 2520}{75} = \frac{5120}{75} = 68.27$$ **Answer:** The combined mean marks is **68.27**

Problem 7: Median (Ungrouped Data)

Question: Find the median of the following ages of employees: 42, 35, 28, 55, 48, 39, 62, 31, 45

Click to reveal solution **Step 1:** Arrange data in ascending order 28, 31, 35, 39, 42, 45, 48, 55, 62 **Step 2:** Count observations n = 9 (odd number) **Step 3:** Find position of median $$\text{Position} = \frac{n + 1}{2} = \frac{9 + 1}{2} = 5^{th} \text{ position}$$ **Step 4:** Identify the median The 5th value in ordered data = 42 **Answer:** The median age is **42 years**

Problem 8: Median (Even Number of Observations)

Question: Find the median of monthly expenditures (in Rs.): 15000, 18000, 22000, 25000, 28000, 32000

Click to reveal solution **Step 1:** Data is already in ascending order 15000, 18000, 22000, 25000, 28000, 32000 **Step 2:** Count observations n = 6 (even number) **Step 3:** For even n, median = average of middle two values $$\text{Positions} = \frac{n}{2} \text{ and } \frac{n}{2} + 1 = 3^{rd} \text{ and } 4^{th}$$ **Step 4:** Calculate median $$\text{Median} = \frac{22000 + 25000}{2} = \frac{47000}{2} = 23500$$ **Answer:** The median expenditure is **Rs. 23,500**

Problem 9: Median (Grouped Data)

Question: Calculate the median from the following distribution:

Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 8 15 12 10
Click to reveal solution **Step 1:** Create cumulative frequency table | Class | f | Cumulative Frequency (cf) | | --------- | ------ | ------------------------- | | 0-10 | 5 | 5 | | 10-20 | 8 | 13 | | 20-30 | 15 | 28 | | 30-40 | 12 | 40 | | 40-50 | 10 | 50 | | **Total** | **50** | | **Step 2:** Find N/2 $$\frac{N}{2} = \frac{50}{2} = 25$$ **Step 3:** Identify median class - cf just greater than 25 is 28 - Median class is **20-30** **Step 4:** Identify values for formula - L (lower limit of median class) = 20 - f (frequency of median class) = 15 - cf (cumulative frequency before median class) = 13 - h (class width) = 10 **Step 5:** Apply median formula $$\text{Median} = L + \frac{\frac{N}{2} - cf}{f} \times h$$ $$\text{Median} = 20 + \frac{25 - 13}{15} \times 10$$ $$= 20 + \frac{12}{15} \times 10 = 20 + 8 = 28$$ **Answer:** The median is **28**

Problem 10: Mode (Ungrouped Data)

Question: Find the mode of the following data representing number of complaints received per day: 5, 3, 7, 3, 8, 5, 3, 9, 5, 3, 7, 3

Click to reveal solution **Step 1:** Arrange and count frequency of each value | Value | Frequency | | ----- | --------- | | 3 | 5 | | 5 | 3 | | 7 | 2 | | 8 | 1 | | 9 | 1 | **Step 2:** Identify the mode The value with highest frequency is 3 (appears 5 times) **Answer:** The mode is **3 complaints per day**

Problem 11: Mode (Grouped Data)

Question: Calculate the mode from the following frequency distribution:

Class 10-20 20-30 30-40 40-50 50-60
Frequency 4 8 12 9 7
Click to reveal solution **Step 1:** Identify modal class The class with highest frequency is 30-40 (f = 12) **Step 2:** Note down required values - L (lower limit of modal class) = 30 - $f_1$ (frequency of modal class) = 12 - $f_0$ (frequency of class before modal class) = 8 - $f_2$ (frequency of class after modal class) = 9 - h (class width) = 10 **Step 3:** Apply mode formula $$\text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$ **Step 4:** Substitute values $$\text{Mode} = 30 + \frac{12 - 8}{2(12) - 8 - 9} \times 10$$ $$= 30 + \frac{4}{24 - 17} \times 10 = 30 + \frac{4}{7} \times 10$$ $$= 30 + 5.71 = 35.71$$ **Answer:** The mode is **35.71**

Section B: Measures of Dispersion

Problem 12: Range

Question: Calculate the range from the following data on response times (in minutes): 12, 18, 25, 8, 32, 15, 28, 10

Click to reveal solution **Step 1:** Identify maximum and minimum values - Maximum value = 32 - Minimum value = 8 **Step 2:** Calculate range $$\text{Range} = \text{Maximum} - \text{Minimum} = 32 - 8 = 24$$ **Answer:** The range is **24 minutes**

Problem 13: Mean Deviation from Mean

Question: Calculate the mean deviation from mean for the following data: 4, 8, 12, 16, 20

Click to reveal solution **Step 1:** Calculate the mean $$\bar{x} = \frac{4 + 8 + 12 + 16 + 20}{5} = \frac{60}{5} = 12$$ **Step 2:** Calculate deviations from mean | x | x - x̄ | | x - x̄ | | | --------- | ----- | ------ | ----- | --- | | 4 | -8 | 8 | | 8 | -4 | 4 | | 12 | 0 | 0 | | 16 | 4 | 4 | | 20 | 8 | 8 | | **Total** | | **24** | **Step 3:** Calculate mean deviation $$MD = \frac{\sum|x - \bar{x}|}{n} = \frac{24}{5} = 4.8$$ **Answer:** The mean deviation from mean is **4.8**

Problem 14: Variance and Standard Deviation (Ungrouped Data)

Question: Calculate the variance and standard deviation for: 5, 10, 15, 20, 25

Click to reveal solution **Step 1:** Calculate the mean $$\bar{x} = \frac{5 + 10 + 15 + 20 + 25}{5} = \frac{75}{5} = 15$$ **Step 2:** Create calculation table | x | x - x̄ | (x - x̄)² | | --------- | ----- | -------- | | 5 | -10 | 100 | | 10 | -5 | 25 | | 15 | 0 | 0 | | 20 | 5 | 25 | | 25 | 10 | 100 | | **Total** | **0** | **250** | **Step 3:** Calculate variance (population) $$\sigma^2 = \frac{\sum(x - \bar{x})^2}{n} = \frac{250}{5} = 50$$ **Step 4:** Calculate standard deviation $$\sigma = \sqrt{50} = 7.07$$ **For sample variance and SD:** $$s^2 = \frac{\sum(x - \bar{x})^2}{n-1} = \frac{250}{4} = 62.5$$ $$s = \sqrt{62.5} = 7.91$$ **Answer:** - Population variance = **50**, SD = **7.07** - Sample variance = **62.5**, SD = **7.91**

Problem 15: Variance and SD (Grouped Data)

Question: Calculate the variance and standard deviation for the following distribution:

Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 8 15 12 10
Click to reveal solution **Step 1:** Create calculation table | Class | x | f | fx | x² | fx² | | --------- | --- | ------ | -------- | ---- | --------- | | 0-10 | 5 | 5 | 25 | 25 | 125 | | 10-20 | 15 | 8 | 120 | 225 | 1800 | | 20-30 | 25 | 15 | 375 | 625 | 9375 | | 30-40 | 35 | 12 | 420 | 1225 | 14700 | | 40-50 | 45 | 10 | 450 | 2025 | 20250 | | **Total** | | **50** | **1390** | | **46250** | **Step 2:** Calculate mean $$\bar{x} = \frac{\sum fx}{\sum f} = \frac{1390}{50} = 27.8$$ **Step 3:** Calculate variance using formula $$\sigma^2 = \frac{\sum fx^2}{\sum f} - \left(\frac{\sum fx}{\sum f}\right)^2$$ $$\sigma^2 = \frac{46250}{50} - (27.8)^2 = 925 - 772.84 = 152.16$$ **Step 4:** Calculate standard deviation $$\sigma = \sqrt{152.16} = 12.34$$ **Answer:** Variance = **152.16**, Standard Deviation = **12.34**

Problem 16: Standard Deviation (Short-cut Method)

Question: Using assumed mean A = 25, calculate SD for the data in Problem 15.

Click to reveal solution **Step 1:** Create table with deviations | Class | x | f | d = x - 25 | fd | d² | fd² | | --------- | --- | ------ | ---------- | ------- | --- | -------- | | 0-10 | 5 | 5 | -20 | -100 | 400 | 2000 | | 10-20 | 15 | 8 | -10 | -80 | 100 | 800 | | 20-30 | 25 | 15 | 0 | 0 | 0 | 0 | | 30-40 | 35 | 12 | 10 | 120 | 100 | 1200 | | 40-50 | 45 | 10 | 20 | 200 | 400 | 4000 | | **Total** | | **50** | | **140** | | **8000** | **Step 2:** Apply short-cut formula $$\sigma = \sqrt{\frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2}$$ $$\sigma = \sqrt{\frac{8000}{50} - \left(\frac{140}{50}\right)^2}$$ $$= \sqrt{160 - 7.84} = \sqrt{152.16} = 12.34$$ **Answer:** Standard Deviation = **12.34** (same as direct method)

Problem 17: Coefficient of Variation

Question: Two departments have the following salary data:

Department Mean Salary (Rs.) SD (Rs.)
A 45,000 8,000
B 62,000 9,500

Which department has more relative variability?

Click to reveal solution **Step 1:** Calculate CV for Department A $$CV_A = \frac{\sigma}{\bar{x}} \times 100 = \frac{8000}{45000} \times 100 = 17.78\%$$ **Step 2:** Calculate CV for Department B $$CV_B = \frac{\sigma}{\bar{x}} \times 100 = \frac{9500}{62000} \times 100 = 15.32\%$$ **Step 3:** Compare - CV_A = 17.78% - CV_B = 15.32% **Answer:** **Department A** has more relative variability (higher CV). Despite Department B having higher absolute SD, Department A's salaries are more variable relative to its mean.

Problem 18: Quartiles and Interquartile Range

Question: Calculate Q₁, Q₂, Q₃, and IQR for: 12, 15, 18, 20, 22, 25, 28, 30, 35, 40, 45

Click to reveal solution **Step 1:** Data is already arranged (n = 11) **Step 2:** Calculate Q₁ position $$Q_1 = \frac{n+1}{4}^{th} \text{ value} = \frac{12}{4} = 3^{rd} \text{ value}$$ $Q_1 = 18$ **Step 3:** Calculate Q₂ (Median) position $$Q_2 = \frac{n+1}{2}^{th} \text{ value} = \frac{12}{2} = 6^{th} \text{ value}$$ $Q_2 = 25$ **Step 4:** Calculate Q₃ position $$Q_3 = \frac{3(n+1)}{4}^{th} \text{ value} = \frac{36}{4} = 9^{th} \text{ value}$$ $Q_3 = 35$ **Step 5:** Calculate IQR $$IQR = Q_3 - Q_1 = 35 - 18 = 17$$ **Answer:** Q₁ = **18**, Q₂ = **25**, Q₃ = **35**, IQR = **17**

Problem 19: Quartiles (Grouped Data)

Question: Calculate Q₁ and Q₃ for the distribution:

Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 8 15 12 10
Click to reveal solution **Step 1:** Create cumulative frequency table | Class | f | cf | | ----- | --- | --- | | 0-10 | 5 | 5 | | 10-20 | 8 | 13 | | 20-30 | 15 | 28 | | 30-40 | 12 | 40 | | 40-50 | 10 | 50 | **Step 2:** Calculate Q₁ - Position = N/4 = 50/4 = 12.5 - Q₁ class = 10-20 (cf = 13 > 12.5) $$Q_1 = L + \frac{\frac{N}{4} - cf}{f} \times h = 10 + \frac{12.5 - 5}{8} \times 10$$ $$= 10 + \frac{7.5}{8} \times 10 = 10 + 9.375 = 19.375$$ **Step 3:** Calculate Q₃ - Position = 3N/4 = 37.5 - Q₃ class = 30-40 (cf = 40 > 37.5) $$Q_3 = 30 + \frac{37.5 - 28}{12} \times 10 = 30 + \frac{9.5}{12} \times 10$$ $$= 30 + 7.917 = 37.917$$ **Answer:** Q₁ = **19.375**, Q₃ = **37.917**, IQR = **18.542**

Problem 20: Combined Standard Deviation

Question: Two groups have the following statistics:

Group n Mean SD
A 50 40 5
B 60 50 6

Find the combined mean and standard deviation.

Click to reveal solution **Step 1:** Calculate combined mean $$\bar{x}_c = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{50 \times 40 + 60 \times 50}{110}$$ $$= \frac{2000 + 3000}{110} = \frac{5000}{110} = 45.45$$ **Step 2:** Calculate deviations from combined mean $$d_1 = \bar{x}_1 - \bar{x}_c = 40 - 45.45 = -5.45$$ $$d_2 = \bar{x}_2 - \bar{x}_c = 50 - 45.45 = 4.55$$ **Step 3:** Apply combined SD formula $$\sigma_c = \sqrt{\frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}}$$ $$= \sqrt{\frac{50(25 + 29.70) + 60(36 + 20.70)}{110}}$$ $$= \sqrt{\frac{50(54.70) + 60(56.70)}{110}} = \sqrt{\frac{2735 + 3402}{110}}$$ $$= \sqrt{\frac{6137}{110}} = \sqrt{55.79} = 7.47$$ **Answer:** Combined Mean = **45.45**, Combined SD = **7.47**

Practice Problems (Try These)

  1. Calculate mean, median, and mode for: 15, 20, 25, 20, 30, 35, 20, 40

  2. Find variance and SD for: 100, 110, 120, 130, 140

  3. The following shows ages: | Age | 20-30 | 30-40 | 40-50 | 50-60 | |—–|——-|——-|——-|——-| | f | 10 | 25 | 18 | 7 |

    Calculate mean, median, mode, and SD.

  4. Two batches have: Batch 1 (n=30, mean=72, SD=8), Batch 2 (n=45, mean=68, SD=10). Find combined statistics.

  5. Calculate coefficient of variation if mean = 50 and SD = 10.

Summary of Key Formulas

Measure Ungrouped Grouped
Mean $\bar{x} = \frac{\sum x}{n}$ $\bar{x} = \frac{\sum fx}{\sum f}$
Median Middle value $L + \frac{N/2 - cf}{f} \times h$
Mode Most frequent $L + \frac{f_1-f_0}{2f_1-f_0-f_2} \times h$
Variance $\frac{\sum(x-\bar{x})^2}{n}$ $\frac{\sum fx^2}{\sum f} - \bar{x}^2$
SD $\sqrt{\text{Variance}}$ $\sqrt{\text{Variance}}$
CV $\frac{\sigma}{\bar{x}} \times 100\%$ $\frac{\sigma}{\bar{x}} \times 100\%$