NCERT Solutions Class 10 Chapter 13 Statistics Exercise 13.1 (original) (raw)

Last Updated : 23 Jul, 2025

The Statistics is an essential branch of mathematics that deals with the collection, organization, analysis, interpretation and presentation of data. In Class 10, Chapter 13 of the NCERT Mathematics textbook covers the various statistical methods such as the mean, median, mode and graphical representation of the data. Exercise 13.1 focuses on the problems that involve calculating measures of the central tendency including mean, median and mode based on the provided datasets.

Statistics

The Statistics is a field that helps in understanding and interpreting data by using the mathematical concepts. It involves collecting, analyzing and summarizing large sets of information to the make meaningful conclusions. The methods used in statistics allow us to the represent data in an organized way and find patterns or trends that help in decision-making processes.

**Question 1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

**Which method did you use for finding the mean, and why?

**Solution:

**Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

**Class Mark = (Upper class Limit + Lower Class Limit)/2

**Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

**Step 3: Now we will apply the general formula to calculate the mean

\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

**Now, Let's see the detailed solution:

**Now, after creating this table we will be able to find the mean very easily -

\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

= 16

= **8.1

**Hence, we come to the conclusion that the number of plants per house is 8.1. Since the numeral value of frequency(fi) and the class mark(xi) is small so we use DIRECT METHOD to find the mean number of plants per house.

**Question 2. Consider the following distribution of daily wages of 50 workers of a factory.

**Find the mean daily wages of the workers of the factory by using an appropriate method.

**Solution:

**Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

**Class Mark = (Upper class Limit + Lower Class Limit)/2

**Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 150, and the class interval is h = 20.

ui = (xi – A)/h

=> ui = (xi – 150)/20

**Step 3: Now we will apply the Assumed Mean Formula to calculate the mean

\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

**Now, Let's see the detailed solution:

**So, the formula to find out the mean is:

**Mean = \large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

= 150 + (20 × -12/50)

= 150 – 4.8

= **145.20

**Thus, mean daily wage of the workers = Rs. 145.20.

**Question 3. The following distribution shows the daily pocket allowance of children of a locality. **The mean pocket allowance is Rs 18. Find the missing frequency f.

**Solution:

**Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

**Class Mark = (Upper class Limit + Lower Class Limit)/2

**Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency. As a certain frequency is missing and we have an odd number of class intervals hence, we will assume the middle-Class Mark as our Assumed Mean(A).

**Step 3: Now we will apply the general formula to calculate the mean

\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

**Now, Let's see the detailed solution:

**The mean formula is

**Mean = \large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

= (752 + 20f)/(44 + f)

**Now substitute the values and equate to find the missing frequency (f)

⇒ 18 = (752 + 20f)/(44 + f)

⇒ 18(44 + f) = (752 + 20f)

⇒ 792 + 18f = 752 + 20f

⇒ 792 + 18f = 752 + 20f

⇒ 792 – 752 = 20f – 18f

⇒ 40 = 2f

⇒ f = 20

**So, the missing frequency, f = 20.

**Question 4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarized as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

**Solution:

**Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

**Class Mark = (Upper class Limit + Lower Class Limit)/2

**Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 75.5 and class size is h = 3.

di = (xi – A)

=> di = (xi – 75.5)

**Step 3: Now we will apply the Assumed Mean Formula to calculate the mean

\large \overline{x}=A+\frac{\sum f_id_i}{\sum f_i}

**Now, Let's see the detailed solution:

**Mean = \large \overline{x}=A+\frac{\sum f_id_i}{\sum f_i}

= 75.5 + (12/30)

= 75.5 + 2/5

= 75.5 + 0.4

= **75.9

**Therefore, the mean heartbeats per minute for these women is 75.9

**Question 5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution of mangoes according to the number of boxes.

**Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

**Solution:

**Step 1: In the above table we find that the class intervals are not continuous and hence to make them a continuous set of data we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals is 1. Then find the Mid Point by using the formula

**Class Mark = (Upper class Limit + Lower Class Limit)/2

**Step 2: In this case, let us assume the mean value, A = 57 and class size is h = 3.

**Step 3: Since the frequency values are big, hence we are using the STEP-DEVIATION METHOD.

\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

**Now, Lets see the detailed solution:

**Mean = \large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

= 57 + 3 * (25/400)

= 57 + 0.1875

= **57.19

**Therefore, the mean number of mangoes kept in a packing box is 57.19

**Question 6. The table below shows the daily expenditure on the **food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

**Solution:

**Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

**Class Mark = (Upper class Limit + Lower Class Limit)/2

**Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 225 and class size is h = 50.

di = (xi – A)

=> di = (xi – 225)

ui = (xi – A)/h

=> ui = (xi – 225)/50

**Step 3: Now we will apply the Step Deviation Formula to calculate the mean

\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

**Now, Let's see the detailed solution:

**Mean = \large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

= 225 + 50 (-7/25)

= 225 - 14

= 211

**Therefore, the mean daily expenditure on food is ₹211

**Question 7. To find out the concentration of SO 2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

**Find the mean concentration of SO 2 in the air.

**Solution:

**Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

**Class Mark = (Upper class Limit + Lower Class Limit)/2

**Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

**Step 3: Now we will apply the general formula to calculate the mean

\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

**Now, Let's see the detailed solution:

The formula to find out the mean is

**Mean = \large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

= 2.96/30

= **0.099 ppm

**Therefore, the mean concentration of SO 2 in the air is 0.099 ppm.

**Question 8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

**Solution:

**Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

**Class Mark = (Upper class Limit + Lower Class Limit)/2

**Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

**Step 3: Now we will apply the general formula to calculate the mean

\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

**Now, Let's see the detailed solution:

The mean formula is,

**Mean = \large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

= 499/40

= **12.48 days

**Therefore, the mean number of days a student was absent = 12.48.

**Question 9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

**Solution:

**Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

**Class Mark = (Upper class Limit + Lower Class Limit)/2

**Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 70 and class size is h = 10.

di = (xi – A)

=> di = (xi – 70)

ui = (xi – A)/h

=> ui = (xi – 70)/10

**Step 3: Now we will apply the Step Deviation Formula to calculate the mean

\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

**Now, Let's see the detailed solution:

So,

**Mean = \large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

= 70 + (-2/35) × 10

= **69.42

**Therefore, the mean literacy rate = 69.42%.

Read more :

• Statistics in Maths
• Top 50 Plus Interview Questions for Statistics with Answers 2024

Conclusion

Exercise 13.1 of Chapter 13 is focused on building the foundational knowledge of the calculating measures of the central tendency such as mean, median and mode. These calculations allow students to the better understand how data can be summarized to the represent the overall characteristics of the dataset. Mastering these concepts will aid in higher-level statistical studies and help in the practical situations where data analysis is required.