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Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.5 | Set 2
  • Last Updated : 05 Mar, 2021

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

Classes0-5050-100100-150150-200200-250250-300300-350
Frequency2356531

Solution:

Let mean (A) = 175

Classes

Class Marks

(x)

Frequency



(f)

c.f.

di = x – A

A = 175

fi * di
0-502522-150-300
50-1007535-100-300
100-150125510-50-250
150-200175-A61600
200-25022552150250
250-300275324100300
300-350325125150150
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.

Number of cars0-1010-2020-3030-4040-5050-6060-7070-80
Frequency71413122011158

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:

Monthly consumption(in units)65-8585-105105-125125-145145-165165-185185-205
No. of consumers4513201484

Solution:

Let mean (A) = 135

Monthly consumptionClass Marks (x)No. of consumers (f)c.f.d = x – Af.d
65-857544-60-240
85-1059559-40-200
105-1251151322-20-260
125-145135204200
145-165155145620280
165-18517586440320
185-20519546860240
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:

Number of letters1-44-77-1010-1313-1616-19
Number of surnames630401644

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 lettersClass Marks(x)No. of surnames (f)c.f.d = x- Af.d.
1-42.566-6-36
4-75.53036-3-90
7-108.5-A407600
10-1311.51692348
13-1614.5496624
16-1917.54100936
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:

Class0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120120 – 140
Frequency81012653

Solution:

Class intervalMid valueFrequency (f)fxCumulative frequency
0 – 20106606
20 – 4030824017
40 – 60501050024
60 – 80701284036
80 – 10090654042
100 – 120110555047
120 – 140130339050
  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:

ExpenditureFrequencyExpenditureFrequency
1000 – 1500243000 – 350030
1500 – 2000403500 – 400022
2000 – 2500334000 – 450016
2500 – 3000284500 – 50007

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 fiXidi = xi – 2750Uifiui
1000 – 1500241250-1500-3-72
1500 – 2000401750-1000-2-80
2000 – 2500332250-500-1-33
2500 – 3000282750000
3000 – 3500303250500130
3500 – 40002237501000244
4000 – 45001642501500348
4500 – 5000747502000428
Total200   -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.

Runs scoredNo. of batsmenRuns scoredNo. of batsmen
3000 – 400047000 – 80006
4000 – 5000188000 – 90003
5000 – 600099000 – 100001
6000 – 7000710000 – 110001

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:

Area of land (in hectares):1 – 33 – 55 – 77 – 99 – 1111 – 13
Number of families204580554012

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:

Income in (in Rs)Number of families
0 – 50008
5000 – 1000026
10000 – 1500041
15000 – 2000016
20000 – 250003
25000 – 300003
30000 – 350002
35000 – 400001

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.

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