# NCERT Solutions for Class 10 Maths Chapter 14 - Statistics

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NCERT Solutions for Class 10 Maths Chapter 14 - Statistics

Mathematics Chapter Statistics is an advanced version of the concept learnt in grade 9. Students are expected to know how to group the ungrouped data and analyse it. This chapter makes a student understand the process of applying concepts to grouped data. Class 10 Maths Chapter Statistics will also discuss the concept of cumulative frequency, the cumulative frequency distribution and how to draw cumulative frequency curves called ogives. Mathematics Chapter Statistics also enhances the knowledge of analysing the data and finding out more information regarding it. This topic holds importance not just in CBSE Class 10, but also in competitive exams. A lot of questions are asked around the concepts and understanding of topics like mean, median and mode. Apart from academic advantages, this chapter also has a lot of practical advantages as it helps students develop the logic of grouping data and analysing it too. Given below are the key topics discussed in this chapter:

Mean of Grouped Data

Mean of Grouped Data using Direct Method

Mean of Grouped Data using Assumed Mean Method.

Mean of Grouped Data using Step-deviation

Mode of Grouped Data

Median of Grouped Data

Graphical Representation of Cumulative Frequency Distribution

Empirical relationship between the three measures of central tendency

Maths Chapter Statistics Chapter 14 - Statistics Exercise Ex. 14.1

Learn the appropriate method to find out the mean of the given grouped data using the Direct method, Assumed Mean method and Step Deviation respectively in the first exercise of Mathematics Chapter Statistics. These questions fall in the 3-mark question category. Students are advised to refer to CBSE Class 10 Maths most important questions, CBSE Class 10 Maths past year papers and CBSE Class 10 Maths sample papers to understand the kinds of questions asked from Mathematics Chapter Statistics. Attempt all the questions from this exercise and check the methods and calculations using Statistics NCERT Solutions. Statistics NCERT Solutions will be the best option for students to boost their confidence in understanding all the topics of this chapter. CBSE Class 10 Maths undoubt is a feature where our subject experts will cater to all the doubts raised by students.

Maths Chapter Statistics Chapter 14 - Statistics Exercise Ex. 14.2

Grouped data has many terminologies such as modal class and mode. Modal class is a class with the maximum frequency and mode is the value inside the modal class that has the highest frequency. Mathematics Chapter Statistics mentions the formula to calculate the mode and TopperLearning’s Statistics NCERT Solutions help you with the method and calculations. Go through all the questions from this exercise and tally your solutions using CBSE Class 10 Maths textbook solutions. Our study material and NCERT solutions provide you with stepwise solutions including the calculations assuring complete clarity. However, if there are any doubts, they can be resolved by our subject experts using CBSE Class 10 Maths doubt solver. CBSE Class 10 Maths video lessons and CBSE Class 10 Maths revision notes are means of adding an extra level of clarity for the students for Mathematics Chapter Statistics. Students will feel more confident in Mathematics Chapter Statistics after referring to TopperLearning’s study material.

Maths Chapter Statistics Chapter 14 - Statistics Exercise Ex. 14.3

Median is the statistical measure that signifies the middle value in a set of data. Discover the process of calculating the median in a grouped set of data. The complexities of the method are simplified with TopperLearning's detailed stepwise Statistics NCERT solutions. These solutions are prepared keeping CBSE guidelines into consideration. Students are advised to go through CBSE Class 10 Maths video lessons, CBSE Class 10 Maths sample papers, CBSE Class 10 Maths most important questions, CBSE Class 10 Maths past year papers and CBSE Class 10 Maths revision notes for best results.

Maths Chapter Statistics Chapter 14 - Statistics Exercise Ex. 14.4

Now that you have studied all the three measures of central tendency, let us discuss which measure would be best suited for a particular requirement in this exercise of Mathematics Chapter Statistics. Learn about the empirical relationship between mean, median and mode. Using this relationship and the methods used to find mean, median and mode respectively, solve the problem from this exercise of Maths Chapter Statistics and compare your solutions with CBSE Class 10 Maths textbook Solutions. TopperLearning’s study material will assure to boost your preparation assuring sure shot success in the board exams. Keep practising and feel free to consult our subject experts for assistance in all your queries at CBSE Class 10 Maths doubt solver.

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## Chapter 14 - Statistics Exercise Ex. 14.1

Solution 1

Let us find class marks (xi) for each interval by using the relation.

Now we may compute xi and fixias following

 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, we may observe that

So, the mean number of plants per house is 8.1.
We have used here direct method as values of class marks (xi) and fi are small.

Solution 2

Let us find class mark for each interval by using the relation.

Class size (h) of this data = 20

Now taking 550 as assured mean (a) we may calculate di, ui and fiuias following.

 Daily wages (in Rs) Number of workers (fi) xi di = xi - 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 we may observe that

So mean daily wages of the workers of the factory is Rs. 545.2.

Solution 3
We may find class mark (xi) for each interval by using the relation.

Given that mean pocket allowance = Rs.18
Now taking 18 as assured mean (a) we may calculate di and fidi as following.
 Daily pocket allowance (in Rs.) Number of children fi 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 may obtain

Hence the missing frequency f is 20.
Solution 4
We may find class mark of each interval (xi) by using the relation.

Class size h of this data = 3
Now taking 75.5 as assumed mean (a) we may calculate di, ui, fiui as following.

 Number of heart beats per minute Number of women fi 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

Now we may observe from table that

So mean heart beats per minute for these women are 75.9 beats per minute.
Solution 5
 Number of mangoes Number of boxes fi 50 - 52 15 53 - 55 110 56 - 58 135 59 - 61 115 62 - 64 25

We may observe that class intervals are not continuous. There is a gap of 1 between two class intervals. So we have to add to upper class limit and subtract from lower class limit of each interval.
And class mark (xi) may be obtained by using the relation

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculate di, ui, fiui as following -

 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

Now we may observe that

Clearly, mean number of mangoes kept in a packing box is 57.19.
We have chosen step deviation method here as values of fi, di are big and also there is a common multiple between all di.

Solution 6

We may calculate class mark (xi) for each interval by using the relation

Class size = 50

Now taking 225 as assumed mean (a) we may calculate di, ui, fiui as following

 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

Now we may observe that -

Solution 7
 Concentration of SO2 (in ppm) Frequency 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

Solution 8
We may find class mark of each interval by using the relation

Now taking 16 as assumed mean (a) we may calculate di and fidi as following

 Number of days Number of students fi xi di = xi - 16 fidi 0 - 6 11 3 -13 -143 6 -10 10 8 -8 -80 10 - 14 7 12 -4 -28 14 - 20 4 16 0 0 20 - 28 4 24 8 32 28 - 38 3 33 17 51 38 - 40 1 39 23 23 Total 40 -145

Now we may observe that

So, mean number of days is 12.38 days, for which a student was absent.

Solution 9

We may find class marks by using the relation

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) we may calculate di, ui, and fiui as following

 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

Now we may observe that

So, the mean literacy rate is 69.43%.

## Chapter 14 - Statistics Exercise Ex. 14.2

Solution 1

We may compute class marks (xi) as per the relation

Now taking 30 as assumed mean (a) we may calculate di and fidi as following.

 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 may observe that

Clearly, 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.
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 (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

Clearly mode is 36.8.It represents that maximum number of patients admitted in hospital were of 36.8 years.
Solution 2
From the data given as above we may observe that maximum class frequency is 61 belonging to class interval 60 - 80.
So, 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

So, modal lifetime of electrical components is 65.625 hours.
Solution 3

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

Now we may find classmark as

Class size (h) of give data = 500
Now taking 2750 as assumed mean (a) we may calculate di, ui and fiui as following

 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

Now from the table, we may observe that

So, mean monthly expenditure was Rs.2662.50.

Solution 4

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 (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 states/U.T have a teacher  student ratio as 30.6

Now we may find class marks by using the relation

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

 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

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

Solution 5
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 (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

So mode of given data is 4608.7 runs.
Solution 6
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 (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 12
Frequency (f2) of class succeeding modal class = 11
Class size = 10

So mode of this data is 44.7 cars.

## Chapter 14 - Statistics Exercise Ex. 14.3

Solution 1
We may find class marks by using the relation

Taking 135 as assumed mean (a) we may find di, ui, fiui, according to step deviation method as following

 Monthly consumption (in units) Number of consumers (f i) xi class mark di = xi - 135 fiui 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 may observe that

Now from table it is clear that 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

We know that
3 median = mode + 2 mean
= 135.76 + 2 (137.058)
= 135.76 + 274.116
= 409.876
Median = 136.625
So median, mode, mean of given data is 136.625, 135.76, 137.05 respectively.

Solution 2
We may find cumulative frequency for the given data as following

 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 - 5 y 40 + x + y 50 - 60 5 45 + x + y Total (n) 60

It is clear that n = 60

45 + x + y = 60

x + y = 15        (1)
Median of data is given as 28.5 which lies in interval 20 - 30.
So, 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 values of x and y are 8 and 7 respectively.

Solution 3

Here the class width is not the same. There is no need to adjust the frequencies according to class intervals. Now given frequency table is of less than type represented with upper-class limits. As policies were given only to persons having age 18 years onwards but less than 60 years, we can define class intervals with their respective cumulative frequency as below

 Age (in years) Number of policyholders (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)

Now from the table we may observe that n = 100.

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

So, the median age is 35.76 years.

Solution 4

The given data is not having continuous class intervals. We can observe that the difference between the two class intervals is 1. So, we have to add and subtract

to upper-class limits and lower class limits.
Now continuous class intervals with respective cumulative frequencies can be represented as below

 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, we may observe that 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

So, the median length of leaves is 146.75 mm.

Solution 5

We can find cumulative frequencies with their respective class intervals as below -

 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

Now we may observe that 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

So, the median lifetime of lamps is 3406.98 hours.

Solution 6

We can find cumulative frequencies with their respective class intervals as below

 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

Now we may observe that cumulative frequency just greater than is 76 belonging to the 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

Now we can find class marks of given class intervals by using relation

Taking 11.5 as assumed mean (a) we can find di, ui and fiui according to step deviation method as below.

 Number of letters Number of surnames xi xi - a 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

fiui = - 106
fi = 100

We know that
3 median = mode + 2 mean
3(8.05) = mode + 2(8.32)
24.15 - 16.64 = mode
7.51 = mode
So, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.51.

Solution 7

We may find cumulative frequencies with their respective class intervals as below

 Weight (in kg) No. of students Cumulative frequency (c.f) 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 = 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 class preceeding the median class = 13
Class size (h) = 5

So, the median weight is 56.67 kg.

## Chapter 14 - Statistics Exercise Ex. 14.4

Solution 1
We can find frequency distribution table of less than type as following -

 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

Now taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we can draw its ogive as following -

Solution 2
The given cumulative frequency distributions of less than type is -

 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

Now taking upper class limits on x-axis and their respective cumulative frequency on y-axis we may draw its ogive as following -

>

Now mark the point A whose ordinate is 17.5 its x-coordinate is 46.5. So median of this data is 46.5.
We may observe that difference between two consecutive upper class limits is 2. Now we may obtain class marks with their respective frequencies 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

Now 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

So median of this data is 46.5
Hence, value of median is verified.

Solution 3
We can obtain cumulative frequency distribution of more than type as following -

 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

Now taking lower class limits on x-axis and their respective cumulative frequencies on y-axis we can obtain its ogive as following.

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