Class 10 NCERT Solutions - Chapter 14 Statistics - Exercise 14.1

Statistics is a branch of mathematics that deals with data collection, analysis, interpretation, and presentation. It plays a crucial role in various fields helping individuals and organizations make informed decisions based on the data. In the Class 10 NCERT curriculum, Chapter 14 on Statistics provides a fundamental understanding of statistical methods and their applications. Exercise 14.1 in this chapter focuses on the basic concepts and calculations involving statistics including mean, median, and mode which are essential for interpreting and understanding data effectively.

Importance of Statistics in Class 10

• Foundation for Advanced Studies: Understanding basic statistical concepts lays the groundwork for the more advanced topics in mathematics and data science.
• Practical Applications: Statistics helps students make sense of real-world data which is valuable in everyday decision-making.
• Skill Development: Learning statistics enhances analytical skills and critical thinking which are useful in various academic and professional fields.
• Preparation for Future Studies: The Mastery of Statistical Techniques prepares students for higher studies in subjects like economics, social sciences, and data analysis.

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 (x_i) 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

Now, Let's see the detailed solution: 

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

= 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.

u_i = (x_i – A)/h 

=> u_i = (xi – 150)/20

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

Now, Let's see the detailed solution: 

So, the formula to find out the mean is:

Mean = 

           = 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

Now, Let's see the detailed solution: 

The mean formula is

Mean = 

          = (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.

d_i = (x_i – A) 

=> d_i = (x_i – 75.5)

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

Now, Let's see the detailed solution: 

Mean = 

= 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.

Now, Lets see the detailed solution: 

Mean = 

= 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.

d_i = (x_i – A) 

=> d_i = (x_i – 225)

u_i = (x_i – A)/h

=> u_i = (x_i – 225)/50

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

Now, Let's see the detailed solution: 

Mean = 

= 225 + 50 (-7/25)

= 225 - 14

= 211

Therefore, the mean daily expenditure on food is ₹211

Question 7. To find out the concentration of SO₂ 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 SO₂ 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

Now, Let's see the detailed solution: 

The formula to find out the mean is

Mean = 

= 2.96/30

= 0.099 ppm

Therefore, the mean concentration of SO₂ 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

Now, Let's see the detailed solution: 

The mean formula is,

Mean = 

= 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.

d_i = (x_i – A) 

=> d_i = (x_i – 70)

u_i = (x_i – A)/h

=> u_i = (x_i – 70)/10

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

Now, Let's see the detailed solution: 

So, 

Mean = 

= 70 + (-2/35) × 10

= 69.42

Therefore, the mean literacy rate = 69.42%.

Read More:

• Statistics in Maths
• Introduction of Statistics and its Types