NCERT Solutions For Class 10 Math Chapter – 14 Exercise – 14.2

NCERT Solutions For Class 10 Math Chapter – 14 Exercise – 14.2

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:

To find out the modal class, let us the consider the class interval with high frequency

Here, the greatest frequency = 23, so the modal class = 35 – 45,

l = 35,

class width (h) = 10,

f_m = 23,

f₁ = 21 and f₂ = 14

The formula to find the mode is

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

Mode = 35+(20/11) = 35+1.8

Mode = 36.8 year

So the mode of the given data = 36.8 year

Calculation of Mean:

First find the midpoint using the formula, x_i= (upper limit +lower limit)/2

Class Interval	Frequency (f_i)	Mid-point (x_i)	f_ix_i
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 f_i = 80		Sum f_ix_i = 2830

The mean formula is

Mean = x̄ = ∑f_ix_i /∑f_i

= 2830/80

= 35.37 years

Therefore, the mean of the given data = 35.37 years

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

Lifetime (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:

From the given data the modal class is 60–80.

l = 60,

The frequencies are:

f_m = 61, f₁ = 52, f₂ = 38 and h = 20

The formula to find the mode is

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values in the formula, we get

Mode =60+[(61-52)/(122-52-38)]×20

Mode = 60+((9 x 20)/32)

Mode = 60+(45/8) = 60+ 5.625

Therefore, modal lifetime of the components = 65.625 hours.

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	Number of families
1000-1500	24
1500-2000	40
2000-2500	33
2500-3000	28
3000-3500	30
3500-4000	22
4000-4500	16
4500-5000	7

Solution:

Given data:

Modal class = 1500-2000,

l = 1500,

Frequencies:

f_m = 40 f₁ = 24, f₂ = 33 and

h = 500

Mode formula:

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values in the formula, we get

Mode =1500+[(40-24)/(80-24-33)]×500

Mode = 1500+((16×500)/23)

Mode = 1500+(8000/23) = 1500 + 347.83

Therefore, modal monthly expenditure of the families = Rupees 1847.83

Calculation for mean:

First find the midpoint using the formula, x_i=(upper limit +lower limit)/2

Let us assume 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

The formula to calculate the mean,

Mean = x̄ = a +(∑f_iu_i /∑f_i)×h

Substitute the values in the given formula

= 2750+(-35/200)×500

= 2750-87.50

= 2662.50

So, the mean monthly expenditure of the families = Rupees 2662.50

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

No of Students per teacher	Number of states / U.T
15-20	3
20-25	8
25-30	9
30-35	10
35-40	3
40-45	0
45-50	0
50-55	2

Solution:

Given data:

Modal class = 30 – 35,

l = 30,

Class width (h) = 5,

f_m = 10, f₁ = 9 and f₂ = 3

Mode Formula:

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values in the given formula

Mode = 30+((10-9)/(20-9-3))×5

Mode = 30+(5/8) = 30+0.625

Mode = 30.625

Therefore, the mode of the given data = 30.625

Calculation of mean:

Find the midpoint using the formula, x_i=(upper limit +lower limit)/2

Class Interval	Frequency (f_i)	Mid-point (x_i)	f_ix_i
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 f_i = 35		Sum f_ix_i = 1022.5

Mean = x̄ = ∑f_ix_i /∑f_i

= 1022.5/35

= 29.2

Therefore, mean = 29.2

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

Run Scored	Number of Batsman
3000-4000	4
4000-5000	18
5000-6000	9
6000-7000	7
7000-8000	6
8000-9000	3
9000-10000	1
10000-11000	1

Find the mode of the data.

Solution:

Given data:

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000,

f_m = 18, f₁ = 4 and f₂ = 9

Mode Formula:

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values

Mode = 4000+((18-4)/(36-4-9))×1000

Mode = 4000+(14000/23) = 4000+608.695

Mode = 4608.695

Mode = 4608.7 (approximately)

Thus, the mode of the given data is 4608.7 runs

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: