Class 10 RD Sharma Solutions Chapter 7 Statistics Exercise 7.5 | Set 2 (original) (raw)

Last Updated : 5 Mar, 2021

Question 11. Find the mean, median, and mode of the following data:

Solution:

Let mean (A) = 175

Classes	Class Marks (x)	Frequency (f)	c.f.	di = x - A A = 175	fi * di
0-50	25	2	2	-150	-300
50-100	75	3	5	-100	-300
100-150	125	5	10	-50	-250
150-200	175-A	6	16	0	0
200-250	225	5	21	50	250
250-300	275	3	24	100	300
300-350	325	1	25	150	150
Total		25			-150

Find Median:

Here N = 25, 5/3 = 25/2 = 12.5 or 13, it lies in the class interval = 50-200.

l = 150, F = 10, f = 6, h = 50

Using median formula, we get

Median = l + \frac{\frac{N}{2}-F}{f} * h \\ = 150 + \frac{12.5-10}{6} * 50 \\ = 150 + \frac{125}{6}

= 150 + 20.83

= 170.83

Find Mean:

Using mean formula we get

Mean = A + \frac{\sum f d_i}{\sum f} \\ = 175 + \frac{-150}{25}

= 175 - 6

= 169

Find Mode:

Using mode formula we get

Mode = l + \frac{f-f_1}{2f-f_1-f_2} * h \\ = 150 + \frac{6-5}{2*6-5-5} * 50

= 150 + 25

= 175

Question 12. 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:

From the given table we conclude that

Modal class = 40-50 (it has maximum frequency)

Also,

l = 40, f = 20, f1 = 12, f2 = 11 and h = 10

By using mode formula, we get

Mode = l + \frac{f-f_1}{2f-f_1-f_2} * h \\ = 40 + \frac{20-12}{2*20-12-11} * 10 \\ = 40 + \frac{8*10}{40-23} \\ = 40 + \frac{8*10}{17}

= 40 + 4.70

= 44.7

Question 13. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean, and mode of the data and compare them:

Solution:

Let mean (A) = 135

Monthly consumption	Class Marks (x)	No. of consumers (f)	c.f.	d = x - A	f.d
65-85	75	4	4	-60	-240
85-105	95	5	9	-40	-200
105-125	115	13	22	-20	-260
125-145	135	20	42	0	0
145-165	155	14	56	20	280
165-185	175	8	64	40	320
185-205	195	4	68	60	240
Total		68			140

Find Median:

Here, N = 34

N/2 = 34,

Class interval = 25-145

Also,

l = 125, F = 22, f = 20 and h = 20

By using the median formula, we get

Median = l + \frac{\frac{N}{2}-F}{f} * h \\ = 125 + \frac{34-22}{20} * 20

= 125 + 12

= 137 units

Find Mean:

By using the mean formula, we get

Mean = A + \frac{\sum f_id_i}{\sum f_i} \\ = 135 + \frac{140}{68}

= 135 + 2.05

= 137.05 units

Find Mode:

By using the mode formula, we get

Mode = l + \frac{f-f_1}{2f-f_1-f_2} * h \\ = 125 + \frac{20-13}{2*20-14-13} * 10 \\ = 125 + \frac{7*20}{40-27} \\ = 125 + \frac{140}{13}

= 125 + 10.76

= 135.76 units

Question 14. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, And the modal size of the surnames.

Solution:

Let mean (A) = 8.5

Number of letters	Class Marks(x)	No. of surnames (f)	c.f.	d = x- A	f.d.
1-4	2.5	6	6	-6	-36
4-7	5.5	30	36	-3	-90
7-10	8.5-A	40	76	0	0
10-13	11.5	16	92	3	48
13-16	14.5	4	96	6	24
16-19	17.5	4	100	9	36
Total		100			-18

Find Median:

Here, N = 100

So, N/2 = 50

Class interval = 7-10

l = 7, F = 36, f = 40 and h =3

By using the median formula, we get

Median = l + \frac{\frac{N}{2}-F}{f} * h \\ = 7 + \frac{50-36}{40} * 3 \\ = 7 + \frac{14}{40} * 3

= 7 + 1.05

= 8.05

Find Mean:

By using the mean formula, we get

Mean = A + \frac{\sum f_id_i}{\sum f_i} \\ = 8.5 + \frac{-18}{100}

= 8.5 + 0.18

= 8.32

Find Mode:

We have,

N = 100

N/2 = 100/2 = 50

Here, the cumulative frequency is just greater than N/2 = 76,

Hence, the median class = 7 - 10

l = 7, h = 10 - 7 = 3, f = 40, F = 36

By using the mode formula, we get

Mode = l + \frac{f-f_1}{2f-f_1-f_2}\times h

= 7 + \frac{40-30}{2\times40-30-16}\times3

= 7 + 30/34

= 7 + 0.88

= 7.88

Question 15. Find the mean, median, and mode of the following data:

Solution:

Class interval	Mid value	Frequency (f)	fx	Cumulative frequency
0 - 20	10	6	60	6
20 - 40	30	8	240	17
40 - 60	50	10	500	24
60 - 80	70	12	840	36
80 - 100	90	6	540	42
100 - 120	110	5	550	47
120 - 140	130	3	390	50
		N = 50	∑fx = 3120

Find Mean:

By using the mean formula, we get

Mean = \frac{∑fx}{N}=\frac{3120}{50}=62.4

Find Median:

We have,

N = 50

Then, N/2 = 50/2 = 25

Here, the cumulative frequency just greater than N/2 = 36

Hence, the median class = 60 - 80

l = 60, h = 80 - 60 = 20, f = 12, F = 24

By using the median formula, we get

Median = l + \frac{\frac{N}{2}-F}{f}\times h

= 60 + \frac{25-24}{12}\times20

= 60 + 20/12

= 60 + 1.67

= 61.67

Find Mode:

We have,

The maximum frequency = 12

Model class = 60 - 80

l = 60, h = 80 - 60 = 20, f = 12, f1 = 10, f2 = 6

By using the mode formula, we get

Mode = l + \frac{f-f_1}{2f-f_1-f_2}\times h

= 60 + \frac{12-10}{2× 12-10-6}\times 20

= 60 + 40/8

= 65

Question 16. 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:

From the given table we conclude that

The maximum class frequency = 40

So, modal class = 1500 - 2000

l = 1500, f = 40, h = 500, f1 = 24, f2 = 33

By using the mode formula, we get

Mode = l + \frac{f-f_1}{2f-f_1-f_2}\times h

= 1500 + \frac{40-24}{2× 40-24-33}\times500

= 1500 + \frac{16}{80-57}× 500

= 1500 + 347.826

= 1847.826 ≈ 1847.83

Hence, the modal monthly expenditure = Rs. 1847.83

Now we will find class marks as

Class mark = \frac{upperclasslimit-lowerclasslimit}{2}

Class size (h) of given data = 500

Let mean(a) = 2750, now we are going to calculate diui as follows:

Expenditure (In Rs)	Number of families fi	Xi	di = xi - 2750	Ui	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
Total	200				-35

From the table we conclude that

∑fi = 200

∑fidi = -35

Mean \overline{x} = a + \frac{\sum f_id_i}{\sum f_i}\times h

= 2750 + \frac{-35}{200}\times500

= 2750 - 87.5

= 2662.5

Hence, the mean monthly expenditure = Rs. 2662.5

Question 17. 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:

From the given table we conclude that

The maximum class frequency = 18

So, modal class = 4000 - 5000

and

l = 4000, f = 18, h = 1000, f1 = 4, f2 = 9

By using the mode formula, we get

Mode = l + \frac{f-f_1}{2f-f_1-f_2}× h

= 4000 + \frac{18-4}{2(18)-4-9}\times 1000

= 4000 + (14000/23)

= 4000 + 608.695

= 4608.695

Hence, the mode of given data = 4608.7 runs.

Question 18. The frequency distribution table of agriculture holdings in a village is given below:

Find the modal agriculture holdings of the village.

Solution:

From the given table we conclude that

The maximum class frequency = 80,

So, the modal class = 5-7

and

l = 5, f0 = 45, h = 2, f1 = 80, f2 = 55

By using the mode formula, we get

Mode = l + \left(\frac{f_i-f_0}{2f_1-f_0-f_2}\right)\times h

= 5 + \left(\frac{80-45}{2(80)-45-55}\right)\times 2

= 5 + \frac{35}{60}\times2

= 5 + \frac{35}{30}

= 5 + 1.2

= 6.2

So, the modal agricultural holdings of the village = 6.2 hectares.

Question 19. The monthly income of 100 families are given as below:

Calculate the modal income.

Solution:

From the given table we conclude that

The maximum class frequency = 41,

So, modal class = 10000-15000.

Here, l = 10000, f1 = 41, f0 = 26, f2 = 16 and h = 5000

Therefore, by using the mode formula, we get

Mode = l + \left(\frac{f_i-f_0}{2f_1-f_0-f_2}\right)\times h

= 10000 + \left(\frac{41-26}{2\times41-26-16}\right)\times 5000

= 10000 + \left(\frac{15}{82-42}\right)\times 5000

= 10000 + \left(\frac{15}{40}\right)\times 5000

= 10000 + 15 × 125

= 10000 + 1875

= 11875

So, the modal income = Rs. 11875.