Here is a NCERT Solutions for Class 10 Maths Chapter 14 Statistics. This solutions covers all questions of Chapter 14 Statistics Class 10 as per CBSE Board guidelines from the latest NCERT book for class 10 maths. Following topics and sub-topics in Chapter 14 Statistics are covered.
14.1 Introduction,
14.2 Mean of Grouped Data,
14.3 Mode of Grouped Data,
14.4 Median of Grouped Data,
14.5 Graphical Representation of Cumulative Frequency Distribution,
14.6 Summary.

## Statistics NCERT Solutions – Class 10 Maths

### NCERT Solutions for Class 10 Maths Chapter 14

Exercise 14.1 : Solutions of Questions on Page Number : 270

Q1 : 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?
To find the class mark (xi) for each interval, the following relation is used.
Class mark (xi) = xi and fixi can be calculated as follows.

 Number of Plants Number of house (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.

Q2 : Consider the following distribution of daily wages of 50 worker of a factory.

 Daily wages (in Rs) 100- 120 120- 140 140 – 1 60 160 – 180 180 – 200 Number of workers 12 14 8 6 10

Find the mean daily wages of the workers of the factory by using an appropriate method.
To find the class mark for each interval, the following relation is used. Class size (h) of this data = 20
Taking 150 as assured mean (a), di, ui, and fiui can be calculated as follows.

 Daily wages(in Rs) Number of workers (fi) xi di =xi– 150  fiui 100 – 120 12 110 – 40 – 2 – 24 120 – 140 14 130 – 20 – 1 – 14 140 – 160 8 150 0 0 0 160 – 180 6 170 20 1 6 180 – 200 10 190 40 2 20 Total 50 – 12

From the table, it can be observed that  Therefore, the mean daily wage of the workers of the factory is Rs 145.20.

Q3 : 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

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 childrenfi Class mark xi di =xi– 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 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.

Q4 : 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

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), di, ui, fiui are calculated as follows.

 Number of heart beats per minute Number of womenfi xi di =xi-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.

Q5 : 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?

 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), di, ui, fiui 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.

Q6 : 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.
To find the class mark (xi) for each interval, the following relation is used. Class size = 50
Taking 225 as assumed mean (a), di, ui, fiui 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.

Q7 : 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.
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), di, ui, fiui are calculated as follows.

 Concentration of SO2 (in ppm) Frequency fi Class mark xi di =xi– 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.

Q8 : 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

To find the class mark of each interval, the following relation is used. Taking 17 as assumed mean(a), di, ui, fiui 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.

Q9 : 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

To find the class marks, the following relation is used. Class size (h) for this data = 10
Taking 70 as assumed mean (a), di, ui, fiui 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 : Solutions of Questions on Page Number : 275

Q1 : 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.
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 patientsfi Class markxi 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.

Q2 : 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.
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.

Q3 : 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

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), di, ui, fiui are 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 Q4 : 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

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), di, ui, 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. Therefore, mean of the data is 29.2
It reprasnt that on avrage, teacher-studant ratio was 29.2

Q5 : 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

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.

Q6 : 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

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 : Solutions of Questions on Page Number : 287

Q1 : 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

To find the class marks, the following relation is used. Taking 135 as assumed mean (a), di, ui, fiui are calculated according to step deviation method as follows.

 Monthly consumption (in units) Number of consumers (fi) xi class mark di=xi-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 cla

Q2 : 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

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 n = 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.

Q3 : 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

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.

Q4 : 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)
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.

Q5 : 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.
The cumulative 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.

Q6 : 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.
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), di, ui, 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 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
∑fµi=-106
∑fi=100  Q7 : 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

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.

Exercise 14.4 : Solutions of Questions on Page Number : 293

Q1 : 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 give.
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. Q2 : 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.
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 Q3 : 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. 