--- title: "NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2" url: https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions/exercise-13-2 dateModified: 2026-10-07T15:56:30+00:00 --- # NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 Chapter 13: Statistics. Every question from Exercise 13.2, 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 13.2 ### Question 1 *3 marks · Short answer* The following table shows the ages of the patients admitted in a hospital during a year: Age (in years): 5-15, 15-25, 25-35, 35-45, 45-55, 55-65 Number of patients: 6, 11, 21, 23, 14, 5 Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. **Solution** 1. The maximum class frequency is 23, so the modal class is $35 - 45$, with $l = 35$, $h = 10$, $f_1 = 23$, $f_0 = 21$, and $f_2 = 14$. 2. Using the mode formula $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$, we get $\text{Mode} = 35 + \left(\frac{23 - 21}{2(23) - 21 - 14}\right) \times 10 = 35 + \frac{20}{11} = 36.8$ years. 3. For the mean, the class marks $x_i$ for $5-15, 15-25, 25-35, 35-45, 45-55, 55-65$ are $10, 20, 30, 40, 50, 60$. 4. Using the assumed mean $a = 40$ and $h = 10$, $u_i = \frac{x_i - 40}{10}$ gives $u_i$ values $-3, -2, -1, 0, 1, 2$, and $f_i u_i$ values $-18, -22, -21, 0, 14, 10$, with $\sum f_i u_i = -37$ and $\sum f_i = 80$. 5. Using the step-deviation method $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 40 + \left(\frac{-37}{80}\right) \times 10 = 35.37$ years. 6. Interpretation: The maximum number of patients admitted are of age $36.8$ years, whereas the average age of the patients admitted is $35.37$ years. **Answer:** Modal age = 36.8 years, Mean age = 35.37 years > Common mistake: Taking incorrect lower limit $l$ or confusing preceding and succeeding frequencies in the mode formula. ### Question 2 *3 marks · Short answer* The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : Lifetimes (in hours): 0-20, 20-40, 40-60, 60-80, 80-100, 100-120 Frequency: 10, 35, 52, 61, 38, 29 Determine the modal lifetimes of the components. **Solution** 1. The maximum class frequency is 61, corresponding to the class interval 60 - 80, so the modal class is 60 - 80. 2. Identify the formula parameters: lower limit $l = 60$, class size $h = 20$, frequency of modal class $f_1 = 61$, frequency of preceding class $f_0 = 52$, and frequency of succeeding class $f_2 = 38$. 3. Substitute the values into the mode formula: $\text{Mode} = 60 + \left(\frac{61 - 52}{2(61) - 52 - 38}\right) \times 20$. 4. Calculate the result: $\text{Mode} = 60 + \frac{9}{122 - 90} \times 20 = 60 + \frac{180}{32} = 60 + 5.625 = 65.625$ hours. **Answer:** 65.625 hours > Common mistake: Incorrect identification of $f_0$ as the frequency after the modal class instead of before. ### Question 3 *3 marks · Short answer* The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure : Expenditure (in ₹): 1000-1500, 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000 Number of families: 24, 40, 33, 28, 30, 22, 16, 7 **Part (i)** 1. The maximum frequency is 40, which corresponds to the class interval 1500 - 2000, so the modal class is 1500 - 2000. 2. Substitute $l = 1500$, $h = 500$, $f_1 = 40$, $f_0 = 24$, and $f_2 = 33$ into the mode formula. 3. $\text{Mode} = 1500 + \left(\frac{40 - 24}{2(40) - 24 - 33}\right) \times 500 = 1500 + \frac{8000}{23} = 1847.83$. Answer (i): Rs 1847.83 **Part (ii)** 1. Using the step-deviation method with assumed mean $a = 2750$ and class size $h = 500$, calculate $u_i = \frac{x_i - 2750}{500}$ to get $\sum f_i = 200$ and $\sum f_i u_i = -15$. 2. Substitute the values into the mean formula $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$. 3. $\bar{x} = 2750 + \left(\frac{-15}{200}\right) \times 500 = 2750 - 37.5 = 2662.50$. Answer (ii): Rs 2662.50 **Answer:** Modal expenditure = Rs 1847.83, Mean expenditure = Rs 2662.50 > Common mistake: Wrong choice of assumed mean leading to tedious arithmetic. ### Question 4 *3 marks · Short answer* The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. Number of students per teacher: 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55 Number of states/U.T.: 3, 8, 9, 10, 3, 0, 0, 2 **Part (i)** 1. The maximum frequency is 10, corresponding to the class interval $30 - 35$. So, the modal class is $30 - 35$. 2. Here $l = 30$, $h = 5$, $f_1 = 10$, $f_0 = 9$, and $f_2 = 3$. 3. Using the formula, $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h = 30 + \left(\frac{10 - 9}{2(10) - 9 - 3}\right) \times 5$ 4. $\text{Mode} = 30 + \frac{1}{8} \times 5 = 30 + 0.625 = 30.6$ (approx). Answer (i): 30.6 **Part (ii)** 1. Let us choose the assumed mean $a = 32.5$ and class size $h = 5$. 2. Constructing the table with class marks $x_i$: for $15-20$, $x_i = 17.5$, $u_i = -3$, $f_i u_i = -9$; for $20-25$, $x_i = 22.5$, $u_i = -2$, $f_i u_i = -16$; for $25-30$, $x_i = 27.5$, $u_i = -1$, $f_i u_i = -9$; for $30-35$, $x_i = 32.5$, $u_i = 0$, $f_i u_i = 0$; for $35-40$, $x_i = 37.5$, $u_i = 1$, $f_i u_i = 3$; for $40-45$, $x_i = 42.5$, $u_i = 2$, $f_i u_i = 0$; for $45-50$, $x_i = 47.5$, $u_i = 3$, $f_i u_i = 0$; for $50-55$, $x_i = 52.5$, $u_i = 4$, $f_i u_i = 8$. 3. Total frequency $\sum f_i = 3 + 8 + 9 + 10 + 3 + 0 + 0 + 2 = 35$, and $\sum f_i u_i = -9 - 16 - 9 + 0 + 3 + 0 + 0 + 8 = -23$. 4. Using the step-deviation method, $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 32.5 + \left(\frac{-23}{35}\right) \times 5 = 32.5 - \frac{23}{7} = 32.5 - 3.29 = 29.21$. Answer (ii): 29.2 **Answer:** The mode is 30.6 and the mean is 29.2. > Common mistake: Taking the wrong frequency for $f_0$ or $f_2$ while calculating the mode. ### Question 5 *3 marks · Short answer* The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Runs scored: 3000-4000, 4000-5000, 5000-6000, 6000-7000, 7000-8000, 8000-9000, 9000-10000, 10000-11000 Number of batsmen: 4, 18, 9, 7, 6, 3, 1, 1 Find the mode of the data. **Solution** 1. Identify the maximum class frequency as 18, which corresponds to the class interval $4000-5000$. 2. Set the parameters for the mode formula: lower limit $l = 4000$, class size $h = 1000$, $f_1 = 18$, $f_0 = 4$, and $f_2 = 9$. 3. Substitute these values into the mode formula: $\text{Mode} = 4000 + \frac{18-4}{2(18)-4-9} \times 1000 = 4000 + \frac{14}{23} \times 1000 = 4608.7$. **Answer:** 4608.7 runs > Common mistake: Forgetting to add the lower limit $l$ after calculating the fraction part. ### Question 6 *3 marks · Short answer* A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data : Number of cars: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80 Frequency: 7, 14, 13, 12, 20, 11, 15, 8 **Solution** 1. Identify the maximum frequency as 20, which corresponds to the class interval $40-50$. 2. Set the parameters for the mode formula: lower limit $l = 40$, class size $h = 10$, $f_1 = 20$, $f_0 = 12$, and $f_2 = 11$. 3. Substitute these values into the mode formula: $\text{Mode} = 40 + \frac{20-12}{2(20)-12-11} \times 10 = 40 + \frac{8}{17} \times 10 = 44.7$. **Answer:** 44.7 cars > Common mistake: Incorrectly identifying $f_0$ and $f_2$ from the frequency list. ## 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 13.3](https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions/exercise-13-3) 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