14-STATISTICS


Here we are going through the ncert solutions for class 10 maths chapter 14- statistics in the zoo so before going through the ncert solutions make sure to go through the textbook that helps you to understand the solutions more easily

Chapter 14 – Statistics

(ncert solutions for class 10 maths chapter 14)


EXERCISE 14.1 (ncert solutions for class 10 maths chapter 14)

Question 1:

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants 0 − 2 2 − 4 4 − 6 6 − 8 8 − 10 10 − 12 12 − 14
Number of houses 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

ANSWER:

To find the class mark (xi) for each interval, the following relation is used.

Class mark (xi) = 

xand fixi can be calculated as follows.

Number of plants Number of houses

 

(fi)

xi fixi
0 − 2 1 1 1 × 1 = 1
2 − 4 2 3 2 × 3 = 6
4 − 6 1 5 1 × 5 = 5
6 − 8 5 7 5 × 7 = 35
8 − 10 6 9 6 × 9 = 54
10 − 12 2 11 2 ×11 = 22
12 − 14 3 13 3 × 13 = 39
Total 20   162

From the table, it can be observed that

Mean,

Therefore, mean number of plants per house is 8.1.

Here, direct method has been used as the values of class marks (xi) and fi are small.

Question 2:

Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs) 500 − 520 520 − 540 540 − 560 560 − 580 580 − 600
Number of workers 12 14 8 6 10

Find the mean daily wages of the workers of the factory by using an appropriate method.

ANSWER:

To find the class mark for each interval, the following relation is used.

Class size (h) of this data = 20

Taking 550 as assured mean (a), diui, and fiui can be calculated as follows.

Daily wages

 

(in Rs)

Number of workers (fi) xi di = x− 550 fiui
500 − 520 12 510 − 40 − 2 − 24
520 − 540 14 530 − 20 − 1 − 14
540 − 560 8 550 0 0 0
560 − 580 6 570 20 1 6
580 − 600 10 590 40 2 20
Total 50       − 12

From the table, it can be observed that


=550+(−1250)20=550−245=550−4.8=545.2=550+-125020=550-245=550-4.8=545.2

Therefore, the mean daily wage of the workers of the factory is Rs 545.20.

Question 3:

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency f.

Daily pocket allowance (in Rs) 11 − 13 13 − 15 15 −17 17 − 19 19 − 21 21 − 23 23 − 25
Number of workers 7 6 9 13 f 5 4

ANSWER:

To find the class mark (xi) for each interval, the following relation is used.

Given that, mean pocket allowance, 

Taking 18 as assured mean (a), di and fidi are calculated as follows.

Daily pocket allowance

 

(in Rs)

Number of children

 

fi

Class mark xi di = x− 18 fidi
11 −13 7 12 − 6 − 42
13 − 15 6 14 − 4 − 24
15 − 17 9 16 − 2 − 18
17 −19 13 18 0 0
19 − 21 f 20 2 f
21 − 23 5 22 4 20
23 − 25 4 24 6 24
Total     2f − 40

From the table, we obtain

Hence, the missing frequency, f, is 20.

Question 4:                                                          (ncert solutions for class 10 maths chapter 14)

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute 65 − 68 68 − 71 71 −74 74 − 77 77 − 80 80 − 83 83 − 86
Number of women 2 4 3 8 7 4 2

ANSWER:

To find the class mark of each interval (xi), the following relation is used.

Class size, h, of this data = 3

Taking 75.5 as assumed mean (a), diuifiui are calculated as follows.

Number of heart beats per minute Number of women

 

fi

xi di = x− 75.5 fiui
65 − 68 2 66.5 − 9 − 3 − 6
68 − 71 4 69.5 − 6 − 2 − 8
71 − 74 3 72.5 − 3 − 1 − 3
74 − 77 8 75.5 0 0 0
77 − 80 7 78.5 3 1 7
80 − 83 4 81.5 6 2 8
83 − 86 2 84.5 9 3 6
Total 30       4

From the table, we obtain

Therefore, mean hear beats per minute for these women are 75.9 beats per minute.

Question 5:

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes 50 − 52 53 − 55 56 − 58 59 − 61 62 − 64
Number of boxes 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

ANSWER:

Number of mangoes Number of boxes fi
50 − 52 15
53 − 55 110
56 − 58 135
59 − 61 115
62 − 64 25

It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore,  has to be added to the upper class limit and  has to be subtracted from the lower class limit of each interval.

Class mark (xi) can be obtained by using the following relation.

Class size (h) of this data = 3

Taking 57 as assumed mean (a), diuifiui are calculated as follows.

Class interval fi xi di xi − 57 fiui
49.5 − 52.5 15 51 − 6 − 2 − 30
52.5 − 55.5 110 54 − 3 − 1 − 110
55.5 − 58.5 135 57 0 0 0
58.5 − 61.5 115 60 3 1 115
61.5 − 64.5 25 63 6 2 50
Total 400       25

It can be observed that

Mean number of mangoes kept in a packing box is 57.19.

Step deviation method is used here as the values of fi, di are big and also, there is a common multiple between all di.

Question 6:                                                          (ncert solutions for class 10 maths chapter 14)

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs) 100 − 150 150 − 200 200 − 250 250 − 300 300 − 350
Number of households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

ANSWER:

To find the class mark (xi) for each interval, the following relation is used.

Class size = 50

Taking 225 as assumed mean (a), diuifiui are calculated as follows.

Daily expenditure (in Rs) fi xi di xi − 225 fiui
100 − 150 4 125 − 100 − 2 − 8
150 − 200 5 175 − 50 − 1 − 5
200 − 250 12 225 0 0 0
250 − 300 2 275 50 1 2
300 − 350 2 325 100 2 4
Total 25       − 7

From the table, we obtain

Therefore, mean daily expenditure on food is Rs 211.

Question 7:

To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

concentration of SO2 (in ppm) Frequency
0.00 − 0.04 4
0.04 − 0.08 9
0.08 − 0.12 9
0.12 − 0.16 2
0.16 − 0.20 4
0.20 − 0.24 2

Find the mean concentration of SO2 in the air.

ANSWER:

To find the class marks for each interval, the following relation is used.

Class size of this data = 0.04

Taking 0.14 as assumed mean (a), diui, fiui are calculated as follows.

Concentration of SO2 (in ppm) Frequency

 

fi

Class mark

 

xi

di x− 0.14 fiui
0.00 − 0.04 4 0.02 − 0.12 − 3 − 12
0.04 − 0.08 9 0.06 − 0.08 − 2 − 18
0.08 − 0.12 9 0.10 − 0.04 − 1 − 9
0.12 − 0.16 2 0.14 0 0 0
0.16 − 0.20 4 0.18 0.04 1 4
0.20 − 0.24 2 0.22 0.08 2 4
Total 30       − 31

From the table, we obtain

Therefore, mean concentration of SO2 in the air is 0.099 ppm.

Question 8:                                                        (ncert solutions for class 10 maths chapter 14)

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days 0 − 6 6 − 10 10 − 14 14 − 20 20 − 28 28 − 38 38 − 40
Number of students 11 10 7 4 4 3 1

ANSWER:

To find the class mark of each interval, the following relation is used.

Taking 17 as assumed mean (a), di and fidi are calculated as follows.

Number of days Number of students

 

fi

xi di xi − 17 fidi
0 − 6 11 3 − 14 − 154
6 − 10 10 8 − 9 − 90
10 − 14 7 12 − 5 − 35
14 − 20 4 17 0 0
20 − 28 4 24 7 28
28 − 38 3 33 16 48
38 − 40 1 39 22 22
Total 40     − 181

From the table, we obtain

Therefore, the mean number of days is 12.48 days for which a student was absent.

Question 9:                                                    (ncert solutions for class 10 maths chapter 14)

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %) 45 − 55 55 − 65 65 − 75 75 − 85 85 − 95
Number of cities 3 10 11 8 3

ANSWER:

To find the class marks, the following relation is used.

Class size (h) for this data = 10

Taking 70 as assumed mean (a), diui, and fiui are calculated as follows.

Literacy rate (in %) Number of cities

 

fi

xi di xi − 70 fiui
45 − 55 3 50 − 20 − 2 − 6
55 − 65 10 60 − 10 − 1 − 10
65 − 75 11 70 0 0 0
75 − 85 8 80 10 1 8
85 − 95 3 90 20 2 6
Total 35       − 2

From the table, we obtain

Therefore, mean literacy rate is 69.43%.

EXERCISE 14.2 (ncert solutions for class 10 maths chapter 14)

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.

ANSWER:

To find the class marks (xi), the following relation is used.

Taking 30 as assumed mean (a), di and fidiare calculated as follows.

Age (in years) Number of patients

 

fi

Class mark

 

xi

di xi − 30 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 obtain

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

It can be observed that the maximum class frequency is 23 belonging to class interval 35 − 45.

Modal class = 35 − 45

Lower limit (l) of modal class = 35

Frequency (f1) of modal class = 23

Class size (h) = 10

Frequency (f0) of class preceding the modal class = 21

Frequency (f2) of class succeeding the modal class = 14

Mode =

Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.

Question 2:                                                                  (ncert solutions for class 10 maths chapter 14)

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
Frequency 10 35 52 61 38 29

Determine the modal lifetimes of the components.

ANSWER:

From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 − 80.

Therefore, modal class = 60 − 80

Lower class limit (l) of modal class = 60

Frequency (f1) of modal class = 61

Frequency (f0) of class preceding the modal class = 52

Frequency (f2) of class succeeding the modal class = 38

Class size (h) = 20

Therefore, modal lifetime of electrical components is 65.625 hours.

Question 3:                                              (ncert solutions for class 10 maths chapter 14)

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 (in Rs) 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

ANSWER:

It can be observed from the given data that the maximum class frequency is 40, belonging to 1500 − 2000 intervals.

Therefore, modal class = 1500 − 2000

Lower limit (l) of modal class = 1500

Frequency (f1) of modal class = 40

Frequency (f0) of class preceding modal class = 24

Frequency (f2) of class succeeding modal class = 33

Class size (h) = 500

Therefore, modal monthly expenditure was Rs 1847.83.

To find the class mark, the following relation is used.

Class size (h) of the given data = 500

Taking 2750 as assumed mean (a), diui, and fiuiare calculated as follows.

Expenditure (in Rs) Number of families

 

fi

xi di = xi − 2750 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 obtain

Therefore, mean monthly expenditure was Rs 2662.50.

Question 4:                                                                    (ncert solutions for class 10 maths chapter 14)

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.

Number 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

ANSWER:

It can be observed from the given data that the maximum class frequency is 10 belonging to class interval 30 − 35.

Therefore, modal class = 30 − 35

Class size (h) = 5

Lower limit (l) of modal class = 30

Frequency (f1) of modal class = 10

Frequency (f0) of class preceding modal class = 9

Frequency (f2) of class succeeding modal class = 3

It represents that most of the states/U.T have a teacher-student ratio as 30.6.

To find the class marks, the following relation is used.

Taking 32.5 as assumed mean (a), diui, and fiui are calculated as follows.

Number of students per teacher Number of states/U.T

 

(fi)

xi di = xi − 32.5 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

Therefore, mean of the data is 29.2.

It represents that on an average, teacher−student ratio was 29.2.

Question 5:                                                            (ncert solutions for class 10 maths chapter 14)

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored Number of batsmen
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.

ANSWER:

From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000 − 5000.

Therefore, modal class = 4000 − 5000

Lower limit (l) of modal class = 4000

Frequency (f1) of modal class = 18

Frequency (f0) of class preceding modal class = 4

Frequency (f2) of class succeeding modal class = 9

Class size (h) = 1000

Therefore, mode of the given data is 4608.7 runs.

Question 6:                                                      (ncert solutions for class 10 maths chapter 14)

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

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

ANSWER:

From the given data, it can be observed that the maximum class frequency is 20, belonging to 40 − 50 class intervals.

Therefore, modal class = 40 − 50

Lower limit (l) of modal class = 40

Frequency (f1) of modal class = 20

Frequency (f0) of class preceding modal class = 12

Frequency (f2) of class succeeding modal class = 11

Class size = 10

Therefore, mode of this data is 44.7 cars.

EXERCISE 14.3 (ncert solutions for class 10 maths chapter 14)

Question 1:

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) Number of consumers
65 − 85 4
85 − 105 5
105 − 125 13
125 − 145 20
145 − 165 14
165 − 185 8
185 − 205 4

ANSWER:

To find the class marks, the following relation is used.

Taking 135 as assumed mean (a), diui, fiui are calculated according to step deviation method as follows.

Monthly consumption (in units) Number of consumers (i) xi class mark dixi− 135
65 − 85 4 75 − 60 − 3 − 12
85 − 105 5 95 − 40 − 2 − 10
105 − 125 13 115 − 20 − 1 − 13
125 − 145 20 135 0 0 0
145 − 165 14 155 20 1 14
165 − 185 8 175 40 2 16
185 − 205 4 195 60 3 12
Total 68       7

From the table, we obtain

From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145.

Modal class = 125 − 145

Lower limit (l) of modal class = 125

Class size (h) = 20

Frequency (f1) of modal class = 20

Frequency (f0) of class preceding modal class = 13

Frequency (f2) of class succeeding the modal class = 14

To find the median of the given data, cumulative frequency is calculated as follows.

Monthly consumption (in units) Number of consumers Cumulative frequency
65 − 85 4 4
85 − 105 5 4 + 5 = 9
105 − 125 13 9 + 13 = 22
125 − 145 20 22 + 20 = 42
145 − 165 14 42 + 14 = 56
165 − 185 8 56 + 8 = 64
185 − 205 4 64 + 4 = 68

From the table, we obtain

n = 68

Cumulative frequency (cf) just greater than  is 42, belonging to interval 125 − 145.

Therefore, median class = 125 − 145

Lower limit (l) of median class = 125

Class size (h) = 20

Frequency (f) of median class = 20

Cumulative frequency (cf) of class preceding median class = 22

Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.

The three measures are approximately the same in this case.

Question 2:                                            (ncert solutions for class 10 maths chapter 14)

If the median of the distribution is given below is 28.5, find the values of x and y.

Class interval Frequency
0 − 10 5
10 − 20 x
20 − 30 20
30 − 40 15
40 − 50 y
50 − 60 5
Total 60

ANSWER:

The cumulative frequency for the given data is calculated as follows.

Class interval Frequency Cumulative frequency
0 − 10 5 5
10 − 20 x 5+ x
20 − 30 20 25 + x
30 − 40 15 40 + x
40 − 50 y 40+ x + y
50 − 60 5 45 + x + y
Total (n) 60  

From the table, it can be observed that = 60

45 + x + y = 60

x + y = 15 (1)

Median of the data is given as 28.5 which lies in interval 20 − 30.

Therefore, median class = 20 − 30

Lower limit (l) of median class = 20

Cumulative frequency (cf) of class preceding the median class = 5 + x

Frequency (f) of median class = 20

Class size (h) = 10

From equation (1),

8 + y = 15

y = 7

Hence, the values of x and y are 8 and 7 respectively.

Question 3:                                      (ncert solutions for class 10 maths chapter 14)

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Age (in years) Number of policy holders
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89
Below 50 92
Below 55 98
Below 60 100

ANSWER:

Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.

Age (in years) Number of policy holders (fi) Cumulative frequency (cf)
18 − 20 2 2
20 − 25 6 − 2 = 4 6
25 − 30 24 − 6 = 18 24
30 − 35 45 − 24 = 21 45
35 − 40 78 − 45 = 33 78
40 − 45 89 − 78 = 11 89
45 − 50 92 − 89 = 3 92
50 − 55 98 − 92 = 6 98
55 − 60 100 − 98 = 2 100
Total (n)    

From the table, it can be observed that n = 100.

Cumulative frequency (cf) just greater than  is 78, belonging to interval 35 − 40.

Therefore, median class = 35 − 40

Lower limit (l) of median class = 35

Class size (h) = 5

Frequency (f) of median class = 33

Cumulative frequency (cf) of class preceding median class = 45

Therefore, median age is 35.76 years.

Question 4:

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm) Number or leaves fi
118 − 126 3
127 − 135 5
136 − 144 9
145 − 153 12
154 − 162 5
163 − 171 4
172 − 180 2

Find the median length of the leaves.

(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 − 126.5, 126.5 − 135.5… 171.5 − 180.5)

ANSWER:

The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore,   has to be added and subtracted to upper class limits and lower class limits respectively.

Continuous class intervals with respective cumulative frequencies can be represented as follows.

Length (in mm) Number or leaves fi Cumulative frequency
117.5 − 126.5 3 3
126.5 − 135.5 5 3 + 5 = 8
135.5 − 144.5 9 8 + 9 = 17
144.5 − 153.5 12 17 + 12 = 29
153.5 − 162.5 5 29 + 5 = 34
162.5 − 171.5 4 34 + 4 = 38
171.5 − 180.5 2 38 + 2 = 40

From the table, it can be observed that the cumulative frequency just greater than   is 29, belonging to class interval 144.5 − 153.5.

Median class = 144.5 − 153.5

Lower limit (l) of median class = 144.5

Class size (h) = 9

Frequency (f) of median class = 12

Cumulative frequency (cf) of class preceding median class = 17

Median 

Therefore, median length of leaves is 146.75 mm.

Question 5:

Find the following table gives the distribution of the life time of 400 neon lamps:

Life time (in hours) Number of lamps
1500 − 2000 14
2000 − 2500 56
2500 − 3000 60
3000 − 3500 86
3500 − 4000 74
4000 − 4500 62
4500 − 5000 48

Find the median life time of a lamp.

ANSWER:

Thecumulative frequencies with their respective class intervals are as follows.

Life time Number of lamps (fi) Cumulative frequency
1500 − 2000 14 14
2000 − 2500 56 14 + 56 = 70
2500 − 3000 60 70 + 60 = 130
3000 − 3500 86 130 + 86 = 216
3500 − 4000 74 216 + 74 = 290
4000 − 4500 62 290 + 62 = 352
4500 − 5000 48 352 + 48 = 400
Total (n) 400  

It can be observed that the cumulative frequency just greater than   is 216, belonging to class interval 3000 − 3500.

Median class = 3000 − 3500

Lower limit (l) of median class = 3000

Frequency (f) of median class = 86

Cumulative frequency (cf) of class preceding median class = 130

Class size (h) = 500

Median 

= 3406.976

Therefore, median life time of lamps is 3406.98 hours.

Question 6:                                                                (ncert solutions for class 10 maths chapter 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 letters 1 − 4 4 − 7 7 − 10 10 − 13 13 − 16 16 − 19
Number of surnames 6 30 40 6 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.

ANSWER:

The cumulative frequencies with their respective class intervals are as follows.

Number of letters Frequency (fi) Cumulative frequency
1 − 4 6 6
4 − 7 30 30 + 6 = 36
7 − 10 40 36 + 40 = 76
10 − 13 16 76 + 16 = 92
13 − 16 4 92 + 4 = 96
16 − 19 4 96 + 4 = 100
Total (n) 100  

It can be observed that the cumulative frequency just greater than  is 76, belonging to class interval 7 − 10.

Median class = 7 − 10

Lower limit (l) of median class = 7

Cumulative frequency (cf) of class preceding median class = 36

Frequency (f) of median class = 40

Class size (h) = 3

Median 

= 8.05

To find the class marks of the given class intervals, the following relation is used.

Taking 11.5 as assumed mean (a), diui, and fiui are calculated according to step deviation method as follows.

Number of letters Number of surnames

 

fi

xi di = xi− 11.5 fiui
1 − 4 6 2.5 − 9 − 3 − 18
4 − 7 30 5.5 − 6 − 2 − 60
7 − 10 40 8.5 − 3 − 1 − 40
10 − 13 16 11.5 0 0 0
13 − 16 4 14.5 3 1 4
16 − 19 4 17.5 6 2 8
Total 100       − 106

From the table, we obtain

fiui = −106

fi = 100

Mean, 

= 11.5 − 3.18 = 8.32

The data in the given table can be written as

Number of letters Frequency (fi)
1 − 4 6
4 − 7 30
7 − 10 40
10 − 13 16
13 − 16 4
16 − 19 4
Total (n) 100

From the table, it can be observed that the maximum class frequency is 40 belonging to class interval 7 − 10.

Modal class = 7 − 10

Lower limit (l) of modal class = 7

Class size (h) = 3

Frequency (f1) of modal class = 40

Frequency (f0) of class preceding the modal class = 30

Frequency (f2) of class succeeding the modal class = 16

Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88.

Question 7:                                                    (ncert solutions for class 10 maths chapter 14)

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg) 40 − 45 45 − 50 50 − 55 55 − 60 60 − 65 65 − 70 70 − 75
Number of students 2 3 8 6 6 3 2

ANSWER:

The cumulative frequencies with their respective class intervals are as follows.

Weight (in kg) Frequency (fi) Cumulative frequency
40 − 45 2 2
45 − 50 3 2 + 3 = 5
50 − 55 8 5 + 8 = 13
55 − 60 6 13 + 6 = 19
60 − 65 6 19 + 6 = 25
65 − 70 3 25 + 3 = 28
70 − 75 2 28 + 2 = 30
Total (n) 30  

Cumulative frequency just greater than   is 19, belonging to class interval 55 − 60.

Median class = 55 − 60

Lower limit (l) of median class = 55

Frequency (f) of median class = 6

Cumulative frequency (cf) of median class = 13

Class size (h) = 5

Median 

= 56.67

Therefore, median weight is 56.67 kg.

MISCELLANEOUS EXERCISE

Question 1:

The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rs) 100 − 120 120 − 140 140 − 160 160 − 180 180 − 200
Number of workers 12 14 8 6 10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

ANSWER:

The frequency distribution table of less than type is as follows.

Daily income (in Rs)

 

(upper class limits)

Cumulative frequency
Less than 120 12
Less than 140 12 + 14 = 26
Less than 160 26 + 8 = 34
Less than 180 34 + 6 = 40
Less than 200 40 + 10 = 50

Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows.

Question 2:                                (ncert solutions for class 10 maths chapter 14)

During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg) Number of students
Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.

ANSWER:

The given cumulative frequency distributions of less than type are

Weight (in kg)

 

upper class limits

Number of students

 

(cumulative frequency)

Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35

Taking upper class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows.

Here, n = 35

So,   = 17.5

Mark the point A whose ordinate is 17.5 and its x-coordinate is 46.5. Therefore, median of this data is 46.5.

It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below.

Weight (in kg) Frequency (f) Cumulative frequency
Less than 38 0 0
38 − 40 3 − 0 = 3 3
40 − 42 5 − 3 = 2 5
42 − 44 9 − 5 = 4 9
44 − 46 14 − 9 = 5 14
46 − 48 28 − 14 = 14 28
48 − 50 32 − 28 = 4 32
50 − 52 35 − 32 = 3 35
Total (n) 35  

The cumulative frequency just greater than is 28, belonging to class interval 46 − 48.

Median class = 46 − 48

Lower class limit (l) of median class = 46

Frequency (f) of median class = 14

Cumulative frequency (cf) of class preceding median class = 14

Class size (h) = 2

Therefore, median of this data is 46.5.

Hence, the value of median is verified.

Question 3:                                                  (ncert solutions for class 10 maths chapter 14)

The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (in kg/ha) 50 − 55 55 − 60 60 − 65 65 − 70 70 − 75 75 − 80
Number of farms 2 8 12 24 38 16

Change the distribution to a more than type distribution and draw ogive.

ANSWER:

The cumulative frequency distribution of more than type can be obtained as follows.

Production yield

 

(lower class limits)

Cumulative frequency
more than or equal to 50 100
more than or equal to 55 100 − 2 = 98
more than or equal to 60 98 − 8 = 90
more than or equal to 65 90 − 12 = 78
more than or equal to 70 78 − 24 = 54
more than or equal to 75 54 − 38 = 16

Taking the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows.

 


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