NCERT Solutions Class 10 - Chapter 13 Statistics - Exercise 13.2

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 = l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 35+\left[\frac{(23-21)}{(46-21-14)}\right]×10

= 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 = \bar{x} = ∑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 = l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 60+\left[\frac{(61-52)}{(122-52-38)}\right]×20

= 60+\frac{(9 \times 20)}{32}

= 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 = l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 1500+\left[\frac{(40-24)}{(80-24-33)}\right]×500

= 1500+\frac{16×500}{23}

= 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 = \overline{x} = a +\frac{∑f_iu_i}{∑f_i}×h

On substituting the values in the given formula

= 2750+\frac{-35}{200}×500

= 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 = l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 30+\frac{(10-9)}{(20-9-3)}×5

= 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 = \bar{x} = \frac{∑f_ix_i }{∑f_i}

= 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 = l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 4000+\frac{(18-4)}{(36-4-9)}×1000

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 = l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 40+\frac{(20-12)}{(40-12-11)}×10

Mode = 40 + 80/17 = 40 + 4.7 = 44.7

Hence, the mode of the given data is 44.7 cars