SEBA Class 10 Mathematics Chapter 14 Statistics Exercise 14.3 Solutions Complete Guide | Assam Eduverse
Chapter Overview:
Assam Eduverse offers SEBA Class 10 Mathematics Solutions Chapter 14 Statistics Exercise 14.3, covering data handling, cumulative frequency, mean, median, and mode. These SEBA Class 10 Maths Chapter 14 Statistics Exercise 14.3 solutions are solved step-by-step to help students master SEBA Class 10 Mathematics Solutions Chapter 14 Statistics Exercise 14.3 effectively for SEBA and ASSEB Class 10 exams. Every question in these SEBA Class 10 Mathematics Solutions Chapter 14 Statistics Exercise 14.3 is explained clearly, making complex statistical concepts easy to understand and practice confidently.
Chapter 14 – Statistics teaches students how to organize, analyze, and interpret data. The SEBA Class 10 Statistics Exercise 14.3 Step by Step solutions guide students through frequency tables, cumulative frequencies, and statistical calculations with precision. With ASSEB Class 10 Mathematics Chapter 14 Statistics Solutions, learning SEBA Class 10 Mathematics Solutions Chapter 14 Statistics Exercise 14.3 becomes simple, helping students solve exercises efficiently and excel in SEBA and ASSEB exams.
These Class 10 SEBA Maths Chapter 14 Statistics Guide solutions are written in exam-focused, easy-to-understand language for improved performance. By practicing SEBA Class 10 Mathematics Solutions Chapter 14 Statistics Exercise 14.3, students strengthen their skills in mean, median, mode, and cumulative frequency. The SEBA Class 10 Maths Chapter 14 Statistics Exercise 14.3 solutions also follow ASSEB standards, ensuring complete preparation, better problem-solving, and success in Class 10 Mathematics exams using SEBA Class 10 Mathematics Solutions Chapter 14 Statistics Exercise 14.3.
Step-by-Step Solutions for Exercise 14.3 | ASSEB / SEBA Class 10 Chapter 14 Statistics Solutions
Q1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.
| Monthly consumption (in units) | No. of customers (f) |
|---|---|
| 65-85 | 4 |
| 85-105 | 5 |
| 105-125 | 13 |
| 125-145 | 20 |
| 145-165 | 14 |
| 165-185 | 8 |
| 185-205 | 4 |
Solution:
Total frequency \(N = 68\)
Median class = 125-145, \(l = 125\), \(cf = 22\), \(f = 20\), \(h = 20\)
Median formula:
\[
\text{Median} = l + \frac{N/2 – cf}{f} \cdot h
\]
Substitute values:
\[
\text{Median} = 125 + \frac{34-22}{20} \cdot 20 = 125 + 12 = 137
\]
Modal class = 125-145
Frequency of modal class (\(f_1\)) = 20
Frequency of class preceding modal class (\(f_0\)) = 13
Frequency of class succeeding modal class (\(f_2\)) = 14
Class size (\(h\)) = 20
Mode formula:
\[
\text{Mode} = l + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \cdot h
\]
Substitute values:
\[
\text{Mode} = 125 + \frac{20-13}{2\cdot20 – 13 – 14} \cdot 20 = 125 + \frac{7}{13} \cdot 20 = 125 + 10.77 = 135.77
\]
Mean calculation:
| Class Interval | f | xi | di = xi-a | ui = di/h | fiui |
|---|---|---|---|---|---|
| 65-85 | 4 | 75 | -60 | -3 | -12 |
| 85-105 | 5 | 95 | -40 | -2 | -10 |
| 105-125 | 13 | 115 | -20 | -1 | -13 |
| 125-145 | 20 | 135 | 0 | 0 | 0 |
| 145-165 | 14 | 155 | 20 | 1 | 14 |
| 165-185 | 8 | 175 | 40 | 2 | 16 |
| 185-205 | 4 | 195 | 60 | 3 | 12 |
Mean formula:
\[
\bar{x} = a + h \frac{\sum f_i u_i}{\sum f_i} = 135 + 20 \frac{7}{68} = 137.05
\]
Q2. If the median of a distribution given below is 28.5, find the value of x & y.
| Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| Frequency | 5 | x | 20 | 15 | y | 5 |
Solution:
Total frequency \(N = 60\)
| Class Interval | f | cf |
|---|---|---|
| 0-10 | 5 | 5 |
| 10-20 | x | 5 + x |
| 20-30 | 20 | 25 + x |
| 30-40 | 15 | 40 + x |
| 40-50 | y | 40 + x + y |
| 50-60 | 5 | 45 + x + y |
Median = 28.5, Median class = 20-30, \(l = 20\), \(cf = 5+x\), \(f = 20\), \(h = 10\)
Median formula:
\[
\text{Median} = l + \frac{N/2 – cf}{f} \cdot h
\]
Substitute values:
\[
28.5 = 20 + \frac{30 – (5+x)}{20} \cdot 10
\]
\[
8.5 = \frac{25 – x}{2} \Rightarrow x = 8
\]
From total frequency:
\[
N = 60 = 45 + x + y \Rightarrow y = 7
\]
x = 8, y = 7
Q3. The life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age.
| Age (in years) | No. of policy holders | Cumulative frequency |
|---|---|---|
| 15-20 | 2 | 2 |
| 20-25 | 4 | 6 |
| 25-30 | 18 | 24 |
| 30-35 | 21 | 45 |
| 35-40 | 33 | 78 |
| 40-45 | 11 | 89 |
| 45-50 | 3 | 92 |
| 50-55 | 6 | 98 |
| 55-60 | 2 | 100 |
Solution:
Total frequency \(N = 100\), \(N/2 = 50\)
Median class = 35-40, \(l = 35\), \(cf = 45\), \(f = 33\), \(h = 5\)
Median formula:
\[
\text{Median} = l + \frac{N/2 – cf}{f} \cdot h
\]
Substitute values:
\[
\text{Median} = 35 + \frac{50 – 45}{33} \cdot 5 = 35 + \frac{25}{33} \approx 35.76
\]
Q4. The lengths of 40 leaves in a plant are measured to the nearest mm. Find the median length.
| Length (mm) | No. of leaves | Cumulative frequency |
|---|---|---|
| 117.5-126.5 | 3 | 3 |
| 126.5-135.5 | 5 | 8 |
| 135.5-144.5 | 9 | 17 |
| 144.5-153.5 | 12 | 29 |
| 153.5-162.5 | 5 | 34 |
| 162.5-171.5 | 4 | 38 |
| 171.5-180.5 | 2 | 40 |
Solution:
Total frequency \(N = 40\), \(N/2 = 20\)
Median class = 144.5-153.5, \(l = 144.5\), \(cf = 17\), \(f = 12\), \(h = 9\)
Median formula:
\[
\text{Median} = l + \frac{N/2 – cf}{f} \cdot h
\]
Substitute values:
\[
\text{Median} = 144.5 + \frac{20-17}{12} \cdot 9 = 144.5 + \frac{27}{12} = 144.5 + 2.25 = 146.75
\]
Q5. Distribution of lifetime of 400 neon lamps. Find the median lifetime.
| Lifetime (hours) | No. of lamps | Cumulative frequency |
|---|---|---|
| 1500-2000 | 14 | 14 |
| 2000-2500 | 56 | 70 |
| 2500-3000 | 60 | 130 |
| 3000-3500 | 86 | 216 |
| 3500-4000 | 74 | 290 |
| 4000-4500 | 62 | 352 |
| 4500-5000 | 48 | 400 |
Solution:
Total frequency \(N = 400\), \(N/2 = 200\)
Median class = 3000-3500, \(l = 3000\), \(cf = 130\), \(f = 86\), \(h = 500\)
Median formula:
\[
\text{Median} = l + \frac{N/2 – cf}{f} \cdot h
\]
Substitute values:
\[
\text{Median} = 3000 + \frac{200 – 130}{86} \cdot 500 = 3000 + \frac{35000}{86} \approx 3406.98
\]
Q6. Frequency distribution of surnames by number of letters. Find median, mean and mode.
| Number of letters | 1-4 | 4-7 | 7-10 | 10-13 | 13-16 | 16-19 |
|---|---|---|---|---|---|---|
| Frequency | 6 | 30 | 40 | 16 | 4 | 4 |
Solution:
Total frequency \(N = 100\), \(N/2 = 50\)
Median class = 7-10, \(l = 7\), \(cf = 36\), \(f = 40\), \(h = 3\)
Median formula:
\[
\text{Median} = l + \frac{N/2 – cf}{f} \cdot h
\]
Substitute values:
\[
\text{Median} = 7 + \frac{50-36}{40} \cdot 3 = 7 + 1.05 = 8.05
\]
Modal class = 7-10, \(f_1 = 40\), \(f_0 = 30\), \(f_2 = 16\), \(h = 3\)
Mode formula:
\[
\text{Mode} = l + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \cdot h
\]
Substitute values:
\[
\text{Mode} = 7 + \frac{40-30}{2\cdot40 – 30 -16} \cdot 3 = 7 + 0.88 = 7.88
\]
Mean calculation:
| Class Interval | f | xi | fixi |
|---|---|---|---|
| 1-4 | 6 | 2.5 | 15 |
| 4-7 | 30 | 5.5 | 165 |
| 7-10 | 40 | 8.5 | 340 |
| 10-13 | 16 | 11.5 | 184 |
| 13-16 | 4 | 14.5 | 58 |
| 16-19 | 4 | 17.5 | 70 |
Mean formula:
\[
\bar{x} = \frac{\sum fixi}{\sum f} = \frac{832}{100} = 8.32
\]
Q7. The distribution below gives the weights of 30 students of a class. Find the median weight.
| Weight (kg) | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 |
|---|---|---|---|---|---|---|---|
| Frequency | 2 | 3 | 8 | 6 | 6 | 3 | 2 |
Solution:
Total frequency \(N = 30\), \(N/2 = 15\)
Median class = 55-60, \(l = 55\), \(cf = 13\), \(f = 6\), \(h = 5\)
Median formula:
\[
\text{Median} = l + \frac{N/2 – cf}{f} \cdot h
\]
Substitute values:
\[
\text{Median} = 55 + \frac{15-13}{6} \cdot 5 = 55 + 1.667 \approx 56.67
\]
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