NCERT Solutions Class 10 Chapter 13 Statistics Exercise 13.2 (original) (raw)

Last Updated : 24 Apr, 2024

Question 1. 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:

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

l = 35,

Class width = 10, and the frequencies are

fm = 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.

xi = (upper limit + lower limit)/2

**Class Interval **Frequency (f i ) **Mid-point (x i ) **f i x 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 i x i = 2830

Mean = \bar{x} = ∑fixi /∑fi

= 2830/80

= 35.37 years

**Class Interval	**Frequency (f i )	**Mid-point (x i )	**f i x 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 i x i = 2830

Question 2. 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

**Solution:

According to the given question

The modal class is 60 – 80

l = 60, and the frequencies are

fm = 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.

**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:

According to the question

Modal class = 1500-2000,

l = 1500,and the frequencies are

fm = 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.

xi = (upper limit + lower limit)/2

Let us considered a mean, A be 2750

**Class Interval **f i **x i **d i = x i – a **u i = d i /h **f i u i

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

**f i = 200 **f i u i = -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

**Class Interval	**f i	**x i	d i = x i – a**	u i = d** i /h	**f i u i
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
	f i = 200**				f i u i = -35**

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

**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:

According to the question

Modal class = 30 – 35,

_l = 30,

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

fm = 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.

xi = (upper limit + lower limit)/2

**Class Interval **Frequency (f i ) **Mid-point (x i ) **f i x 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 i x i = 1022.5

Mean = \bar{x} = \frac{∑f_ix_i }{∑f_i}

= 1022.5/35

= 29.2

Hence, the mean is 29.2

**Class Interval	**Frequency (f i )	**Mid-point (x i )	**f i x 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 i x i = 1022.5

Question 5. 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:

According to the question

Modal class = 4000 – 5000,

l = 4000,

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

fm = 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:

**Number of cars	**Frequency
**0-10	**7
**10-20	**14
**20-30	**13
**30-40	**12
**40-50	**20
**50-60	**11
**60-70	**15
**70-80	**8

**Solution:

According to the question

Modal class = 40 – 50, l = 40,

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

fm = 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

NCERT Solutions Class 10 Chapter 13 Statistics Exercise 13.2 (original) (raw)

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.

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.

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:

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

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.

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: