--- title: "NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.3" url: https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions/exercise-13-3 dateModified: 2026-10-07T15:56:30+00:00 --- # NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.3 Chapter 13: Statistics. Every question from Exercise 13.3, with full working and the final answer. Free PDF (13 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-13-statistics-2798fa8577.pdf ## EXERCISE 13.3 ### Question 1 *6 marks · Long answer* The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. Monthly consumption (in units): 65-85, 85-105, 105-125, 125-145, 145-165, 165-185, 185-205 Number of consumers: 4, 5, 13, 20, 14, 8, 4 **Part (i)** 1. Construct the frequency distribution table with class intervals, frequency ($f_i$), class mark ($x_i$), $f_i x_i$, cumulative frequency ($cf$), and step-deviation ($u_i$). 2. For class 125-145, $cf = 22$, $f = 20$, $l = 125$, and $h = 20$. 3. Calculate median using Median = $l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h = 125 + \left(\frac{34 - 22}{20}\right) \times 20 = 137$ units. Answer (i): Median = 137 units **Part (ii)** 1. Using the direct method or step-deviation method with assumed mean $a = 135$ and $h = 20$, compute $\sum f_i x_i = 9320$ and $\sum f_i = 68$. 2. Calculate mean using $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{9320}{68} = 137.06$ units. Answer (ii): Mean = 137.06 units **Part (iii)** 1. Identify the modal class with maximum frequency 20, which is 125-145. 2. Use the mode formula: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h = 125 + \left(\frac{20 - 13}{2(20) - 13 - 14}\right) \times 20 = 135.77$ units. 3. Comparing the three measures, we observe that they are approximately equal, indicating that the data is moderately skewed and representative of a normal distribution. Answer (iii): Mode = 135.77 units, and the three measures are approximately equal. **Answer:** Median = 137 units, Mean = 137.06 units, Mode = 135.77 units > Common mistake: Taking incorrect cumulative frequency or wrong lower limit for the median or modal class. ### Question 2 *3 marks · Short answer* If the median of the distribution given below is 28.5, find the values of $x$ and $y$. Class interval: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 Frequency: 5, $x$, 20, 15, $y$, 5 (Total = 60) **Solution** 1. Set up the cumulative frequency table: class intervals with frequencies $5, x, 20, 15, y, 5$ and cumulative frequencies $5, 5+x, 25+x, 40+x, 40+x+y, 45+x+y$. 2. Use the given total frequency $60$ to write $45 + x + y = 60$, which simplifies to $x + y = 15$. 3. Use the given median $28.5$ to identify the median class as $20 - 30$ with $l = 20$, $f = 20$, $cf = 5 + x$, and $h = 10$. 4. Substitute into the median formula: $28.5 = 20 + \left(\frac{30 - (5 + x)}{20}\right) \times 10$. 5. Solve the equation to find $x = 8$, and substitute $x = 8$ into $x + y = 15$ to get $y = 7$. **Answer:** x = 8 and y = 7 > Common mistake: Taking the wrong cumulative frequency for the median class. ### Question 3 *3 marks · Short answer* A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year. Age (in years): Below 20, Below 25, Below 30, Below 35, Below 40, Below 45, Below 50, Below 55, Below 60 Number of policy holders: 2, 6, 24, 45, 78, 89, 92, 98, 100 **Solution** 1. Given cumulative frequency distribution: Below 20: 2, Below 25: 6, Below 30: 24, Below 35: 45, Below 40: 78, Below 45: 89, Below 50: 92, Below 55: 98, Below 60: 100. 2. Convert the given less-than type cumulative frequency distribution into class intervals with frequencies: 20-25: 4, 25-30: 18, 30-35: 21, 35-40: 33, 40-45: 11, 45-50: 3, 50-55: 6, 55-60: 2. 3. Here $n = 100$, so $\frac{n}{2} = 50$, which lies in the median class $35-40$ with lower limit $l = 35$, cumulative frequency of preceding class $\text{cf} = 45$, frequency $f = 33$, and class size $h = 5$. 4. Using the formula $\text{Median} = l + \left(\frac{\frac{n}{2} - \text{cf}}{f}\right) \times h$, we substitute the values to get $\text{Median} = 35 + \left(\frac{50 - 45}{33}\right) \times 5$. 5. Simplifying the expression gives $\text{Median} = 35 + \frac{25}{33} = 35 + 0.76 = 35.76\text{ years}$. **Answer:** 35.76 years > Common mistake: Subtracting cumulative frequencies incorrectly while finding the class frequencies. ### Question 4 *3 marks · Short answer* The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table : Length (in mm): 118-126, 127-135, 136-144, 145-153, 154-162, 163-171, 172-180 Number of leaves: 3, 5, 9, 12, 5, 4, 2 Find the median length of the leaves. (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5-126.5, 126.5-135.5, ..., 171.5-180.5.) **Solution** 1. Convert the discontinuous class intervals into continuous class intervals by subtracting $0.5$ from the lower limits and adding $0.5$ to the upper limits: $117.5 - 126.5, 126.5 - 135.5, \dots, 171.5 - 180.5$. 2. Write the frequencies and compute the cumulative frequencies: $3, 8, 17, 29, 34, 38, 40$. 3. Find $n = 40$, so $\frac{n}{2} = 20$. 4. Identify the median class as $144.5 - 153.5$ with $l = 144.5$, $f = 12$, $cf = 17$, and $h = 9$. 5. Substitute the values in the median formula: $\text{Median} = 144.5 + \left(\frac{20 - 17}{12}\right) \times 9 = 144.5 + 2.25 = 146.75$ mm. **Answer:** 146.75 mm > Common mistake: Applying the median formula directly without converting the classes into continuous intervals. ### Question 5 *3 marks · Short answer* The following table gives the distribution of the life time of 400 neon lamps : Life time (in hours): 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000 Number of lamps: 14, 56, 60, 86, 74, 62, 48 Find the median life time of a lamp. **Solution** 1. Prepare the cumulative frequency table for the given life times: frequencies are $14, 56, 60, 86, 74, 62, 48$ and cumulative frequencies are $14, 70, 130, 216, 290, 352, 400$. 2. Find $n = 400$, so $\frac{n}{2} = 200$. 3. Identify the median class as $3000 - 3500$ with lower limit $l = 3000$, frequency $f = 86$, cumulative frequency of preceding class $cf = 130$, and class size $h = 500$. 4. Substitute the values into the median formula: $\text{Median} = 3000 + \left(\frac{200 - 130}{86}\right) \times 500$. 5. Calculate the final value: $\text{Median} = 3000 + \frac{70 \times 500}{86} = 3000 + 406.98 = 3406.98$ hours. **Answer:** 3406.98 hours > Common mistake: Arithmetic errors in computing the fractional part of the median. ### Question 6 *3 marks · Short answer* 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Number of letters: 1-4, 4-7, 7-10, 10-13, 13-16, 16-19 Number of surnames: 6, 30, 40, 16, 4, 4 Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames. **Part (i)** 1. Find cumulative frequencies for the classes: $6, 36, 76, 92, 96, 100$. 2. Find $n = 100$, so $\frac{n}{2} = 50$. 3. Identify median class as $7 - 10$ with $l = 7$, $f = 40$, $cf = 36$, and $h = 3$, giving $\text{Median} = 7 + \left(\frac{50 - 36}{40}\right) \times 3 = 8.05$ letters. Answer (i): 8.05 letters **Part (ii)** 1. Find class marks $x_i$: $2.5, 5.5, 8.5, 11.5, 14.5, 17.5$. 2. Calculate $f_i x_i$: $15, 165, 340, 184, 58, 70$, giving $\sum f_i x_i = 832$. 3. Compute mean as $\bar{x} = \frac{832}{100} = 8.32$ letters. Answer (ii): 8.32 letters **Part (iii)** 1. Identify modal class as $7 - 10$ with $l = 7$, $f_1 = 40$, $f_0 = 30$, $f_2 = 16$, and $h = 3$. 2. Substitute into the mode formula: $\text{Mode} = 7 + \left(\frac{40 - 30}{2(40) - 30 - 16}\right) \times 3$. 3. Simplify to get $\text{Mode} = 7 + \frac{10}{34} \times 3 = 7.88$ letters. Answer (iii): 7.88 letters **Answer:** Median = 8.05 letters, Mean = 8.32 letters, Mode = 7.88 letters > Common mistake: Using incorrect class marks for intervals with decimal mid-points. ### Question 7 *3 marks · Short answer* The distribution below gives the weights of 30 students of a class. Find the median weight of the students. Weight (in kg): 40-45, 45-50, 50-55, 55-60, 60-65, 65-70, 70-75 Number of students: 2, 3, 8, 6, 6, 3, 2 **Solution** 1. Prepare the cumulative frequency table for the given weight distribution: classes 40-45 ($f=2, cf=2$), 45-50 ($f=3, cf=5$), 50-55 ($f=8, cf=13$), 55-60 ($f=6, cf=19$), 60-65 ($f=6, cf=25$), 65-70 ($f=3, cf=28$), 70-75 ($f=2, cf=30$). 2. Total frequency $n = 30$, so $\frac{n}{2} = 15$. The cumulative frequency just greater than 15 is 19, corresponding to the median class 55-60. 3. Here $l = 55$, $cf = 13$, $f = 6$, and $h = 5$. 4. Substitute the values into the median formula: $\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h = 55 + \left(\frac{15 - 13}{6}\right) \times 5 = 55 + \frac{10}{6} = 56.67$ kg. **Answer:** Median weight = 56.67 kg > Common mistake: Using the wrong cumulative frequency of the preceding class or incorrect class size $h$. ## Related pages - [All Chapter 13 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions) - [Exercise 13.1](https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions/exercise-13-1) - [Exercise 13.2](https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions/exercise-13-2) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student