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

(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.

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.

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.

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 |

Class interval: | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 |

Frequency: | 30 | 45 | 75 | 35 | 25 | 15 |

Class interval: | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 | 45 – 50 | 50 – 55 |

Frequency: | 25 | 34 | 50 | 42 | 29 | 15 |

(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, f_{1} = 12, f_{2 }= 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, f_{1} = 45, f_{2 }= 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, f_{1} = 34, f_{2 }= 42

= 35 + 80/24

= 35 + 3.33

= 38.33

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 |

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, f_{1} = 50, f_{2 }= 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, f_{1} = 54, f_{2 }= 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

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 |

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, f_{1} = 13, f_{2 }= 5

= 50 + 70/22

= 50 + 3.18

= 53.18

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.

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, f_{1} = 118, f_{2} = 127

= 165.5 + 72/39

= 165.5 + 1.85

= 167.35 cm

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.

We may compute class marks (x_{i}) as per the relation

Now taking 30 as assumed mean (a) we may calculate d_{i }and f_{i}d_{i }as follows:

Age (in years) | Number of patients f_{i} |
Class marks x_{i} |
d_{i }= x_{i} – 275 |
f_{i}d_{i} |

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 Σf_{i} = 80, Σf_{i }d_{i} = 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 (f_{1}) of class preceding the modal class = 21 Frequency (f_{2}) 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.

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.

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 (f_{1}) of class preceding the modal class = 52

Frequency (f_{2}) of class succeeding the modal class = 38 Class size (h) = 20 Mode

So, modal lifetime of electrical components is 65.625 hours

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.

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_{1 }= 12, f_{2 }= 8

= 120 + 5

= 125

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 |

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 (f_{1}) of class preceding modal class = 9

Frequency (f_{2}) 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 d_{i}, u_{i}, and f_{i}u_{i }as following

Number of students per teacher | Number of states/ U.T (f_{i}) |
x_{i} |
d_{i }= x_{i} – 32.5 |
U_{i} |
f_{i}u_{i} |

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

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 |

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_{1 }= 5, f_{2 }= 5

= 150 + 50/2

= 150 + 25

= 175

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.

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 (f_{1}) of class preceding modal class = 12

Frequency (f_{2}) 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

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 |

Class interval | Mid value x_{i} |
Frequency f_{i} |
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, f_{1} = 13, f_{2 }= 14

= 125 + 140/13

= 135.77

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.

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_{1 }= 30, f_{2 }= 16

= 7 + 30/34

= 7 + 0.88

= 7.88

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 |

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_{1 }= 10, f_{2 }= 6

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 |

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 (f_{1}) of class preceding modal class = 24

Frequency (f_{2}) 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 d_{i }u_{i }as follows:

Expenditure (in Rs) | Number of families f_{i} |
x_{i} |
d_{i }= x_{i} – 2750 |
U_{i} |
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 |

Total | 200 | -35 |

Now from table may observe that Σf_{i} = 200, Σf_{i}d_{i} = -35

So mean monthly expenditure was Rs. 2662.5

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.

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 (f_{1}) of class preceding modal class = 4 Frequency (f_{2}) 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

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