Class 10 NCERT Solutions- Chapter 14 Statistics - Exercise 14.2

Chapter 14 of Class 10 NCERT Mathematics focuses on Statistics a branch of mathematics that deals with data collection, organization, analysis, and interpretation. In Exercise 14.2, students work on the organizing continuous data into the grouped frequency distributions and learning how to calculate the mean using the different methods including the direct method, assumed mean method and step-deviation method.

Statistics

The Statistics is a crucial field that helps us make sense of large sets of data by the organizing, analyzing and drawing conclusions from them. In this chapter, we will learn how to represent data using the different techniques like frequency distribution tables and measures of central tendency. This allows us to summarize and interpret the data meaningfully. The Statistics is widely applied in the various real-world fields such as the economics, science and social studies to make informed decisions based on the data.

Question 1. The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:

The greatest frequency in the given table is 23, so the modal class = 35 – 45,

l = 35,

Class width = 10, and the frequencies are

f_m = 23, f₁ = 21 and f₂ = 14

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

= 35 + (20/11) = 35 + 1.8

= 36.8

Hence, the mode of the given data is 36.8 year

Now, we find the mean. So for that first we need to find the midpoint.

x_i = (upper limit + lower limit)/2

Class Interval	Frequency (fi)	Mid-point (xi)	fixi
5-15	6	10	60
15-25	11	20	220
25-35	21	30	630
35-45	23	40	920
45-55	14	50	700
55-65	5	60	300
Sum fi = 80	Sum fixi = 2830

Mean =  = ∑f_ix_i /∑f_i

= 2830/80

= 35.37 years

Question 2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.

Solution:

According to the given question

The modal class is 60 – 80

l = 60, and the frequencies are

f_m = 61, f₁ = 52, f₂ = 38 and h = 20

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

= 

= 60 + 45/8 = 60 + 5.625

Hence, the modal lifetime of the components is 65.625 hours.

Question 3. 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:

Solution:

According to the question

Modal class = 1500-2000,

l = 1500,and the frequencies are

f_m = 40 f₁ = 24, f₂ = 33 and

h = 500

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

= 

= 1500 + 8000/23 = 1500 + 347.83

So, the modal monthly expenditure of the families is 1847.83 Rupees 

Now, we find the mean. So for that first we need to find the midpoint.

x_i = (upper limit + lower limit)/2

Let us considered a mean, A be 2750

Class Interval	fi	xi	di = xi – a	ui = di/h	fiui
1000-1500	24	1250	-1500	-3	-72
1500-2000	40	1750	-1000	-2	-80
2000-2500	33	2250	-500	-1	-33
2500-3000	28	2750	0	0	0
3000-3500	30	3250	500	1	30
3500-4000	22	3750	1000	2	44
4000-4500	16	4250	1500	3	48
4500-5000	7	4750	2000	4	28
fi = 200	fiui = -35

Mean = 

On substituting the values in the given formula

= 

= 2750 - 87.50

= 2662.50

Hence, the mean monthly expenditure of the families is  2662.50 Rupees

Question 4. 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

Solution:

According to the question

Modal class = 30 – 35,

l = 30,

Class width (h) = 5, and the frequencies are

f_m = 10, f₁ = 9 and f₂ = 3

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

= 30 + 5/8 = 30 + 0.625

= 30.625

Hence, the mode of the given data is 30.625

Now, we find the mean. So for that first we need to find the midpoint.

x_i = (upper limit + lower limit)/2

Class Interval	Frequency (fi)	Mid-point (xi)	fixi
15-20	3	17.5	52.5
20-25	8	22.5	180.0
25-30	9	27.5	247.5
30-35	10	32.5	325.0
35-40	3	37.5	112.5
40-45	0	42.5	0
45-50	0	47.5	0
50-55	2	52.5	105.5
Sum fi = 35	Sum fixi = 1022.5

Mean = 

= 1022.5/35 

= 29.2

Hence, the mean is 29.2

Question 5. The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.

Find the mode of the data.

Solution:

According to the question

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000, and the frequencies are

f_m = 18, f₁ = 4 and f₂ = 9

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

Mode = 4000 + 14000/23 = 4000 + 608.695

= 4608.695

Hence, the mode of the given data is 4608.7 runs

Question 6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:

Solution:

According to the question

Modal class = 40 – 50, l = 40,

Class width (h) = 10, and the frequencies are

f_m = 20, f₁ = 12 and f₂ = 11

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

Mode = 40 + 80/17 = 40 + 4.7 = 44.7

Hence, the mode of the given data is 44.7 cars

Conclusion

Exercise 14.2 provides students with an understanding of how to calculate the mean for the grouped data using the different methods. The concepts and calculations help students understand the importance of the statistical methods in the summarizing and interpreting data. Learning these techniques is crucial for the various real-life applications making statistics an essential subject for the students.