 # Chapter 7: Statistics Exercise – 7.5

### Question: 1

Find the mode of the following data:

(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15

### Solution:

(i)

 Value (x) 3 4 5 6 7 8 9 Frequency (f) 4 2 5 2 2 1 2

Mode = 5 because it occurs the maximum number of times.

(ii)

 Value (x) 3 4 5 6 7 8 9 Frequency (f) 5 2 4 2 2 1 2

Mode = 3 because it occurs maximum number of times.

(iii)

 Value (x) 8 15 18 19 20 24 25 26 Frequency (f) 1 4 1 1 1 2 1 1

Mode = 15 because it occurs maximum number of times.

### Question: 2

The shirt size worn by a group of 200 persons, who bought the shirt from a store, are as follows:

 Shirt size: 37 38 39 40 41 42 43 44 Number of persons: 15 25 39 41 36 17 15 12

Find the model shirt size worn by the group.

### Solution:

 Shirt size 37 38 39 40 41 42 43 44 Number of persons 15 25 39 41 36 17 15 12

Model shirt size = 40 because it occurs maximum number of times.

### Question: 3

Find the mode of the following distribution. (i)

 Class interval: 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Frequency: 5 8 7 12 28 20 10 10

### (ii)

 Class interval: 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 Frequency: 30 45 75 35 25 15

### (iii)

 Class interval: 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 50 – 55 Frequency: 25 34 50 42 29 15

### Solution:

(i)

 Class interval 0-10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Frequency 5 8 7 12 28 20 10 10

Here the maximum frequency is 28 then the corresponding class 40 – 50 is the modal class

l = 40, h = 50 40 = 10, f = 28, f1 = 12, f= 20 = 40 + 160/ 24

= 40 + 6.67

= 46.67

(ii)

 Class interval 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 Frequency 30 45 75 35 25 15

Here the maximum frequency is 75, then the corresponding class 20 - 25 is the modal class

l = 20, h = 25 - 20 = 5, f = 75, f1 = 45, f= 35 = 20 + 150/70

= 20 + 2.14

= 22.14

(iii)

 Class interval 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 50 – 55 Frequency 25 34 50 42 38 14

Here the maximum frequency is 50 then the corresponding class 35 - 40 is the modal class

l = 35, h = 40 - 35 = 5, f = 50, f1 = 34, f= 42 = 35 + 80/24

= 35 + 3.33

= 38.33

### Question: 4

Compare the modal ages of two groups of students appearing for an entrance test:

 Age in years 16 – 18 18 – 20 20 – 22 22 – 24 24 – 26 Group A 50 78 46 28 23 Group B 54 89 40 25 17

### Solution:

 Age in years 16 – 18 18 – 20 20 – 22 22 – 24 24 – 26 Group A 50 78 46 28 23 Group B 54 89 40 25 17

For Group A Here the maximum frequency is 78, then the corresponding class 18 – 20 is model class l = 18, h = 20 - 18 = 2, f = 78, f1 = 50, f= 46 = 18 + 56/60

= 18 + 0.93

= 18.93 years

For group B Here the maximum frequency is 89, then the corresponding class 18 - 20 is the modal class l = 18, h = 20 - 18 = 2, f = 89, f1 = 54, f= 40 Mode = 18 + 70/84

= 18 + 0.83

= 18.83 years

Hence the modal age for the Group A is higher than that for Group B

### Question: 5

The marks in science of 80 students of class X are given below. Find the mode of the marks obtained by the students in science.

 Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 Frequency 3 5 16 12 13 20 5 4 1 1

### Solution:

 Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 Frequency 3 5 16 12 13 20 5 4 1 1

Here the maximum frequency is 20, then the corresponding class 50 – 60 is the modal class l = 50, h = 60 – 50 = 10, f = 20, f1 = 13, f= 5 = 50 + 70/22

= 50 + 3.18

= 53.18

### Question: 6

The following is the distribution of height of students of a certain class in a city:

 Height (in cm): 160 – 162 163 – 165 166 – 168 169 – 171 172 – 174 No of students: 15 118 142 127 18

Find the average height of maximum number of students.

### Solution:

 Heights(exclusive) 160 – 162 163 – 165 166 – 168 169 – 171 172 – 174 Heights (inclusive) 159.5 – 162.5 162.5 – 165.5 165.5 – 168.5 168.5 – 171.5 171.5 – 174.5 No of students 15 118 142 127 18

Here the maximum frequency is 142, then the corresponding class 165.5 – 168.5 is the modal class l = 165.5, h = 168.5 - 165.5 = 3, f = 142, f1 = 118, f2 = 127 = 165.5 + 72/39

= 165.5 + 1.85

= 167.35 cm

### Question: 7

The following table shows the ages of the patients admitted in a hospital during a year:

 Ages (in years): 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65 No of students: 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:

We may compute class marks (xi) as per the relation Now taking 30 as assumed mean (a) we may calculate dand fidas follows:

 Age (in years) Number of patients fi Class marks xi di = xi – 275 fidi 5 – 15 6 10 - 20 -120 15  25 11 20 - 10 -110 25  35 21 30 0 0 35  45 23 40 10 230 45 - 55 14 50 20 280 55  65 5 60 30 150 Total 80 430

From the table we may observe that Σfi = 80, Σfi di = 430, = 30 + 430/80

= 30 + 5.375

= 35.375

= 35.38

Clearly, mean of this data is 35.38. It represents that on an average the age of a patients admitted to hospital was 35.38 years.

As we may observe that maximum class frequency is 23 belonging to class interval 35 – 45

So, modal class = 35 – 45 Lower limit (l) of modal class = 35 Frequency (f) of modal class = 23

Class size (h) = 10 Frequency (f1) of class preceding the modal class = 21 Frequency (f2) of class succeeding the modal class = 14 Mode Clearly mode is 36.8. It represents that maximum number of patients admitted in hospital were of 36.8 years.

### Question: 8

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

 Lifetimes (in hours): 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120 No of components: 10 35 52 61 38 29

Determine the modal lifetimes of the components.

### Solution:

From the data given as above we may observe that maximum class frequency is 61 belonging to class interval 60 – 80

So, modal class limit (l) of modal class = 60 Frequency (f) of modal class = 61 Frequency (f1) of class preceding the modal class = 52

Frequency (f2) of class succeeding the modal class = 38 Class size (h) = 20 Mode So, modal lifetime of electrical components is 65.625 hours

### Question: 9

The following table gives the daily income of 50 workers of a factory:

 Daily income 100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 Number of workers 12 14 8 6 10

Find the mean, mode and median of the above data.

### Solution:

 Class interval Mid value (x) Frequency (f) fx Cumulative frequency 100 – 120 110 12 1320 12 120 – 140 130 14 1820 26 140 – 160 150 8 1200 34 160 – 180 170 6 1000 40 180 – 200 190 10 1900 50 N = 50 Σfx = 7260 We have, N = 50 Then, N/2 = 50/2 = 25

The cumulative frequency just greater than N/2 is 26, then the median class is 120 - 140 such that l = 120, h = 140 - 120 = 20, f = 14, F = 12 = 120 + 260/14

= 120 + 18.57

= 138.57

Here the maximum frequency is 14, then the corresponding class 120 - 140 is the modal class l = 120, h = 140 - 120 = 20, f = 14, f= 12, f= 8 = 120 + 5

= 125

### Question: 10

The following distribution gives the state-wise teachers-students ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures:

 Number of students per teacher Number of states/ U.T Number of students per teacher Number of states/ U.T 15 – 20 3 35 – 40 3 20 – 25 8 40 – 45 0 25 – 30 9 45 – 50 0 30 – 35 10 50 – 55 2

### Solution:

We may observe from the given data that maximum class frequency is 10 belonging to class interval 30 – 35.

So, modal class = 30 – 35 Class size (h) = 5 Lower limit (l) of modal class = 30 Frequency (f) of modal class = 10 Frequency (f1) of class preceding modal class = 9

Frequency (f2) of class succeeding modal class = 3 Mode = 30.6 It represents that most of states/ U.T have a teacher- student ratio as 30.6 Now we may find class marks by using the relation Class mark = (upper class limit + lower class limit) /2.

Now taking 32.5 as assumed mean (a) we may calculate di, ui, and fiuas following

 Number of students per teacher Number of states/ U.T (fi) xi di = xi – 32.5 Ui fiui 15 – 20 3 17.5 -15 - 3 - 9 20 – 25 8 22.5 -10 - 2 - 16 25 – 30 9 27.5 -5 - 1 - 9 30 – 35 10 32.5 0 0 0 35 – 40 3 37.5 5 1 3 40 – 45 0 42.5 10 2 0 45 – 50 0 47.5 15 3 0 50 – 55 2 52.5 20 4 8 Total 35 -23

Now, So mean of data is 29.2 It represents that on an average teacher-student ratio was 29.2

### Question: 11

Find the mean, median and mode of the following data:

 Classes: 0 – 50 50 – 100 100 – 150 150 – 200 200 – 250 250 – 300 300 – 350 Frequency: 2 3 5 6 5 3 1

### Solution:

 Class interval Mid value (x) Frequency (f) fx Cumulative frequency 0 – 50 35 2 50 2 50 – 100 75 3 225 5 100 – 150 125 5 625 10 150 – 200 175 6 1050 16 200 – 250 225 5 1127 21 250 – 300 275 3 825 24 300 – 350 325 1 325 25 N = 25 Σfx = 4225

### We have, N = 25

Then, N/2

= 25/2

= 12.5

The cumulative frequency just greater than N/2 is 16, then the median class is 150 - 200 such that l = 150, h = 200 - 150 = 50, f = 6, F = 10 = 150 + 125/6

= 150 + 20.83

= 170.83

Here the maximum frequency is 6, then the corresponding class 150 - 200 is the modal class l = 150, h = 200 - 150 = 50, f = 6, f= 5, f= 5 = 150 + 50/2

= 150 + 25

= 175

### Question: 12

A students noted the number of cars pass through a spot on a road for 100 periods each of 3 minute and summarized it in the table given below. Find the mode of the data.

### Solution:

From the given data we may observe that maximum class frequency is 20 belonging to 40 – 50 class intervals.

So, modal class = 40 – 50 Lower limit (l) of modal class = 40 Frequency (f) of modal class = 20 Frequency (f1) of class preceding modal class = 12

Frequency (f2) of class succeeding modal class = 11 Class size = 10 = 40 + 80/17

= 40 + 4.7

= 44.7

So mode of this data is 44.7 cars

### 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: 65 – 85 85 – 105 105  – 125 125 – 145 145 – 165 165 – 185 185 – 205 No of consumers: 4 5 13 20 14 8 4

### Solution:

 Class interval Mid value xi Frequency fi fx Cumulative frequency 65 – 85 75 4 300 4 85 – 105 95 5 475 9 105 – 125 115 13 1495 22 125 – 145 135 20 2700 42 145 – 165 155 14 2170 56 165  185 175 8 1400 64 185  205 195 4 780 68 N = 68 Σfx = 9320 We have, N = 68 N/2 = 68/2 = 34

The cumulative frequency just greater than N/2 is 42 then the median class is 125 - 145 such that l = 125, h = 145 - 125 = 20, f = 20, F = 22 = 125 + 12

= 137

Here the maximum frequency is 20, then the corresponding class 125 - 145 is the modal class l = 125, h = 145 - 125 = 20, f = 20, f1 = 13, f= 14 = 125 + 140/13

= 135.77

### Question: 14

100 surnames were randomly picked up from a local telephone directly and the frequency distribution of the number of letter English alphabets in the surnames was obtained as follows:

 Number of letters: 1 – 4 4 – 7 7 – 10 10 – 13 13 – 16 16  – 19 Number surnames: 6 30 40 16 4 4

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

### Solution:

 Class interval Mid value (x) Frequency (f) fx Cumulative frequency 1 – 4 2.5 6 15 6 4 – 7 5.5 30 165 36 7 – 10 8.5 40 340 76 10 – 13 11.5 16 184 92 13 – 16 14.5 4 58 96 16 – 19 17.5 4 70 100 N = 100 Σfx = 832 We have, N = 100 N/2 = 100/2 = 50

The cumulative frequency just greater than N/2 is 76, then the median class is 7 - 10 such that l = 7, h = 10 - 7 = 3, f = 40, F = 36 = 7 + 1.05

= 8.05

Here the maximum frequency is 40, then the corresponding class 7 - 10 is the modal class l = 7, h = 10 - 7 = 3, f = 40, f= 30, f= 16 = 7 + 30/34

= 7 + 0.88

= 7.88

### Question: 15

Find the mean, median and mode of the following data:

 Class 0 - 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120 120 – 140 Frequency 6 8 10 12 6 5 3

### Solution:

 Class interval Mid value (x) 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 Σfx = 3120 We have, N = 50 Then, N/2 = 50/2 = 25

The cumulative frequency just greater than N/2 is 36, then the median class is 60 - 80 such that l = 60, h = 80 - 60 = 20, f = 12, F = 24 = 60 + 20/12

= 60 + 1.67

= 61.67

Here the maximum frequency is 12, then the corresponding class 60 - 80 is the modal class l = 60, h = 80 - 60 = 20, f = 12, f= 10, f= 6 ### 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:

 Expenditure Frequency Expenditure Frequency 1000 – 1500 24 3000 – 3500 30 1500 – 2000 40 3500 – 4000 22 2000 – 2500 33 4000 – 4500 16 2500 – 3000 28 4500 – 5000 7

### Solution:

We may observe from the given data that maximum class frequency is 40 belonging to 1500 -200 intervals. So, modal class = 1500 -2000

Lower limit (l) of modal class = 1500 Frequency (f) of modal class = 40 Frequency (f1) of class preceding modal class = 24

Frequency (f2) of class succeeding modal class = 33 Class size (h) = 500 = 1500 + 347.826

= 1847.826

1847.83

So modal monthly expenditure was Rs. 1847.83

Now we may find class mark as Class mark Class size (h) of given data = 500 Now taking 2750 as assumed mean (a) we may calculate duas 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

Now from table may observe that Σfi = 200, Σfidi = -35 So mean monthly expenditure was 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 scored No of batsmen Runs scored No of batsmen 3000 – 4000 4 7000 – 8000 6 4000 – 5000 18 8000 – 9000 3 5000 – 6000 9 9000 – 10000 1 6000 – 7000 7 10000 – 11000 1

Find the mode of the data.

### Solution:

From the given data we may observe that maximum class frequency is 18 belonging to class interval 4000 – 5000.

So, modal class = 4000 – 5000 Lower limit (l) of modal class = 4000

Frequency (f) of modal class = 18. Frequency (f1) of class preceding modal class = 4 Frequency (f2) of class succeeding modal class = 9 Class size (h) = 1000. Now = 4000 + 608.695

= 4608.695

So mode of given data is 4608.7 runs ### Course Features

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