NCERT solutions class 10 maths chapter 14 application of statistics is one of the most important topics in Class 10 mathematics syllabus. Students must practice this chapter well to prepare maths more effectively. The class 10 maths chapter 14 application of statistics NCERT solutions is provided here to help students understand the concepts of the chapter in details. Students must practice the NCERT Solutions for class 10 maths chapter 14 thoroughly to score good marks in their class 10 board examination. Check the class 10 maths NCERT Solutions for chapter 14 pdf given below.
The topics covered in the chapter is given below
Section Number  Topic 
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 
NCERT Solutions Class 10 Maths Chapter 14 Exercises
 NCERT Solutions Class 10 Maths Chapter 14 Statistics Types of statistics Exercise 14.1
 NCERT Solutions Class 10 Maths Chapter 14 Statistics Types of statistics Exercise 14.2
 NCERT Solutions Class 10 Maths Chapter 14 Statistics Types of statistics Exercise 14.3
Exercise – 1
Question 1:
A group of pupil has conducted a survey as a as the part of their college program, They collected examined data’s of plants from 20 houses and nearby areas. Find the mean of the collected plants from the given houses.
No. of Plants
Taken from 20 houses 
02 
24 
46

68 
810 
1012 
1214 
Given number of houses 
2 
1 
5 
1 
2 
6 
3 
Name the method you use for finding the mean and why?
Solution:
Class Interval  f_{i}  x_{i}  f_{i} x_{i} 
02  2  1  1 
24  1  3  6 
46  5  5  5 
68  1  7  35 
810  2  9  54 
1012  6  11  22 
1214  3  13  40 
Sum f_{i}=20  Sum f_{i}x_{i}=163 
Mean can be calculated as follows:
\(\bar{x}= \frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{163}{20}=8.15\)
In this issues the \( f_{i} and x_{i}\) are small so the direct method is given to solve the problem.
Question2:
The distributed wages is given for 50 workers working in the factory.
Daily wages given to workers (in Rs)  100120  120140  140160  160180  180200 
Number of wage workers in factory  14  12  6  8  10 
Find the mean of workers of daily wages given by the workers.
Solution:
In this case, value of xi is quite large and hence we should select the assumed mean method.
Class given Interval  f_{i}  x_{i}  d_{i} = x_{i} – a  f_{i} d_{i} 
100120  14  110  40  480 
120140  12  130  20  280 
140160  6  150  0  0 
160180  8  170  20  120 
180200  10

190  40  400

Sum f_{i}=50  Sum f_{i} d_{i} = 240 
Now, mean of deviations of daily wagesis calculated as follows:
\(\bar{d}=\frac{\sum f_{i}d_{i}}{\sum f_{i}}=\frac{240}{50}\)
x = d + a = 150+(4.8)= 145.20
Question 3:
This distribution shows that the money daily pocket given to students in the given area. The mean pocket money is Rs. 18. Find the f which is the missing frequency.
Daily money allowance (in Rs) 
1113 
1315 
1517 
1719 
1921 
2123 
2325 
Number of children 
7 
6 
9 
13 
f 
5 
4 
Solution:
Class interval  f_{i}  x_{i}  f_{i}x_{i} 
1113  7  12  84 
1315  6  14  84 
1517  13  16  144 
1719  9  18  234 
1921  f  20  20f 
2123  4  22  110 
2325  5  24  96 
Sum f_{i}=44+f  Sum f_{i}x_{i}= 752+20f 
We have;
\(\bar{x}= \frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}\)
\(18= \frac{752+20f}{44+ f_{i}}\)
18(44+18f) = 752+20f
792+18f = 752+20f
2f = 40
Missing frequency f= 20
Question4:
The number of 30 women were checked by one doctor and the distributed the heartbeat of them in following distribution table. Find the mean of heart beats of the given per minute for these thirty women, by choosing a suitable following method.
Heartbeat per minute in given number  6568  6871  7174  7477  7780  8083  8386 
Number of 30 women  2  3  4  7  8  2  4 
Solution:
Class Interval  f_{i}  xi  d_{i} = xi – a  f_{i} d_{i} 
6568  2  66.5  9  18 
6871  3  69.5  6  24 
7174  4  72.5  3  9 
7477  7  75.5  0  0 
7780  8  78.5  3  21 
8083  2  81.5  6  24 
8386  4  84.5  9  18 
Sum f_{i}= 30  Sum f_{i} d_{i} = 12 
Now, mean can be calculated as follows:
\(\bar{d}=\frac{\sum f_{i}d_{i}}{\sum f_{i}}=\frac{12}{30}=0.4\)
\(\bar{x}=\bar{d}+a=0.4+75.5=75.9\)
Question 5:
The oranges were packed and were sold by the fruit seller in a market. The numbers of oranges are arranged in different boxes by the fruit seller. Through the number of oranges the number of oranges are distributed in following manner.
No.of oranges  5052  5355  5658  8961  6264 
No. of boxes  15  110  135  115  25 
In the distributed pattern the number of oranges. Find the mean. Choose the method to find the mean:
Solution:
Class interval  f_{i}  x_{i}  d_{i}=xa  f_{i} d_{i} 
5052  110  54  6  90 
5355  15  51  3  330 
5658  135  60  0  0 
5961  25  57  3  345 
6264  115  63  6  150 
Sum f_{i}= 400  Sum f_{i} d_{i}=75 
Mean can be calculated as follows:
\(\bar{d}=\frac{\Sigma f_{i}d_{i}}{\Sigma f_{i}}=\frac{75}{400}=0.1875\)
\(\bar{x}=\bar{d}+a=0.1875+57=57.1875\)
In this case, there are wide variations in \(f_{i}\) and hence assumed mean method is used.
Question 6:
The following distribution shows the food expenditure of the given households from the locality:
Daily food expenditure (in Rs)  50100  100150  150200  200250  250300 
Number of given households  4  5  12  2  2 
Find the mean daily expenditure on food by a suitable method.
Solution:
Class Interval  f_{i}  x_{i}  d_{i}= xi – a  u_{i} = di/h  f_{i}u_{i} 
50100  4  125  100  2  8 
100150  5  175  50  1  5 
150200  12  225  0  0  0 
200250  2  275  50  1  2 
250300  2  325  100  2  4 
Sum f_{i} = 25  Sum f_{i}u_{i} = 7 
Mean can be calculated as follows:
\(\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h\)
\(=225+\frac{7}{25}\times 50 =225\)
Question 7:
To find concentration in sulphur dioxide (in parts per million, i.e. ppm),The parts of data are collected by different areas from cities are given in distributed table:
Concentration in SO_{2} (in ppm)  Frequency of F 
0.000.04  4 
0.040.08  9 
0.080.12  9 
0.120.16  2 
0.160.20  4 
0.200.24  2 
Give the mean of the sulphur dioxide collected from the given cities.
Solution:
Class Interval  f_{i}  x_{i}  f_{i}x_{i} 
0.000.04  9  0.02  0.08 
0.040.08  4  0.06  0.54 
0.080.12  9  0.10  0.90 
0.120.16  2  0.14  0.28 
0.160.20  2  0.18  0.72 
0.200.24  4  0.22  0.44 
Sum f_{i}=30  Sum f_{i}x_{i}=2.96 
Mean can be calculated as follows:
\(\bar{x}=\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{2.96}{30}=0.099 ppm\)
Question 8:
The absentee record of 40 students is handled by the teacher for one complete term. Find the mean of absent days of the given students:
Number of following days  06  610  1014  1420  2028  2838  3840 
Number of following students  10  11  4  7  3  4  1 
Solution:
Class Interval  f_{i}  x_{i}  f_{i}x_{i} 
06  11  3  33 
610  10  8  80 
1014  4  12  84 
1420  7  17  68 
2028  3  24  96 
2838  4  33  99 
3840  1  39  39 
Sum f_{i} = 40  Sum f_{i}x_{i}= 499 
Mean can be calculated as follows:
\(\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}\)= \(\frac{499}{40}\)=12.4
Question 9:
The data of the literacy digit from the 35 cities is given in the following table. Find the mean of the 35 literacy cities.
Literacy rate from 35cities (in %) 
4045 
5055 
6065 
7075 
8085 
Cities in values  10  3  8  11  3 
Solution:
Class Interval  f_{i}  x_{i}  d_{i}=xia  u_{i}=di/h  f_{i}u_{i} 
4045  10  50  20  2  6 
5055  3  60  10  1  10 
6065  8  70  0  0  0 
7075  11  80  10  1  8 
8085  3  90  20  2  6 
Sum f_{i}=35  Sum f_{i}u_{i}= 2 
Mean can be calculated as follows:
\(\bar{x}=a+\frac{\Sigma f_{i}u_{i}}{\Sigma f_{i}}\times h\)
\(= 70+\frac{2}{35}\times 10=69.42\)
EXERCISE – 2
Question 1:
The age of patients admitted in a hospitals are given in the following table for a complete year.
Age (in years)  515  1525  2535  3545  4555  5565 
Number of patients  6  11  21  23  14  5 
Calculate the mode and mean of the data above. Compare and interpret the two measures of central tendency.
Solution:
Modal class = 3545, length = 35, height = 10, \( a_{1} = 23, a_{0} = 21\, and\, a_{3} = 14\)
Therefore, Mode = \(l +\frac{a_{1}a_{0}}{2a_{1}a_{0}a_{2}}\times h\)
= \(35+\left ( \frac{2321}{2\times 232114} \right )\times 10\)
= \(35+\frac{2}{11}\times 10=36.8\)
Calculating Mean
CLASS INTERVAL  a_{i}  z_{i}  a_{i}z_{i} 
515  6  10  60 
1525  11  20  220 
2535  21  30  630 
3545  21  30  920 
4555  14  50  700 
5565  5  60  300 
Sum a_{i} = 80  Sum a_{i}z_{i}= 2830 
\(\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a_{i}}=\frac{2830}{80}=35.37\)
\(\ therefore\)The mode of data displays that the maximum number of patients is in the age group of 26.8, whereasthe average age of all the patients is 35.37.
Question 2:
The data given below provides the information for the observed lifetime (in hours) of 225 electrical components:
Lifetime ( in hours)  0 – 20  20 – 40  40 – 60  60 – 80  80 – 100  100 – 120 
Frequency  10  35  52  61  38  29 
Find the modal lifetimes of the components.
Solution:
Modal class = 6080, l = 60,\( a_{1} = 61, a_{0} = 52, a_{2} = 38\) and h = 20
\(Mode = l + \left ( \frac{a_{1}a_{0}}{2a_{1}a_{0}a_{2}} \right )\times h\)
\(= 60 + \left ( \frac{61 – 52}{2 \times 61 – 52 – 38 } \right ) \times 20\)
\(= 60 + \frac{9}{32} \times 20 = 65.62\)
Question 3:
The total monthly household expenses of 200 families of a town are given in the following table. Calculate the modal monthly expenses of the families. Also, calculate the mean monthly expenses.
Expenses  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 
Solution:
Modal class = 15002000, l = 1500,\( a_{1} = 40, a_{0} = 24, a_{2} = 33\) and h = 500
\(Mode = l + \left ( \frac{a_{1}a_{0}}{2a_{1}a_{0}a_{2}} \right )\times h\)
\(= 1500 + \left ( \frac{40 – 24}{2 \times 40 – 24 – 33} \right ) \times 500\)
\(= 1500 + \left ( \frac{16}{23} \right ) \times 500 = 1847.82\)
Calculation of mean:
Class Interval  a_{i}  z_{i}  d_{i} =zi – b  u_{i}= di/h  a_{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 
a_{i} = 200  a_{i}u_{i} = 35 
\(\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a{i}}\times h\)
\(= 2750 – \frac{35}{200} \times 500\)
= 2750 – 87.5 = 2662.50
Question 4:
The state – wise teacher student ratio in a college in India are given in the following table. Calculate the mean and mode of the given data. Interpret the two measures.
Number of students per teacher  Number of states/UT 
15 – 20  3 
20 – 25  8 
25 – 30  9 
30 – 35  10 
35 – 40  3 
40 – 45  0 
45 – 50  0 
50 – 55  2 
Solution:
Modal class = 3035, l = 30, \(a_{1} = 10, a_{0} = 9, a_{2} = 3\) and h = 5
\(Mode = l +\left ( \frac{a_{1}a_{0}}{2a_{1}a_{0}a_{2}} \right )\times h\)
\(= 30 – \left ( \frac{20 – 9}{2 \times 10 – 9 – 3 } \right ) \times 5\)
\(= 30 – \frac{1}{8} \times 5 = 30.625\)
Calculating mean:
Number of students per teacher  a_{i}  z_{i}  d_{i}= z_{i} –b  u_{i} = u_{i}/h  a_{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 
Sum a_{i} = 35  sum a_{i}u_{i}= 23 
\(\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a_{i}}\times h\)
\(= 30.5 – \frac{23}{35}\times 5\)
\(= 30.5 – \frac{23}{7} = 29.22\)
Therefore, from the mode we know that the maximum number of states has 30 – 35 students per teacher, and from the mean we know that the average ratio of students per teacher is 29.22
Question 5:
Provided table shows the total runs scored by some of the top batsman of the world in oneday cricket matches.
Runs Scored  Number of Batsman 
3000 – 4000  4 
4000 – 5000  18 
5000 – 6000  9 
6000 – 7000  7 
7000 – 8000  6 
8000 – 9000  3 
9000 – 10000  1 
10000 – 11000  1 
Calculate the mode of the given information.
Solution:
Modal class = 40005000, l = 4000, \(a_{1} = 18, a_{0} = 4, a_{2} = 9\) and h = 1000
\(Mode = l + \left ( \frac{a_{1}a_{0}}{2a_{1}a_{0}a_{2}} \right )\times h\)
\(= 4000 + \left ( \frac{18 – 4}{2 \times 18 – 4 9} \right )\times 1000\)
\(= 4000 + \left ( \frac{14}{23} \right )\times 1000 = 4608.70\)
Question 6:
A complete observation of the number of cars on the roads for 100 periods of 3 minutes are summarized in the following table. Calculate 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 
Solution:
Modal class = 40 – 50, l = 40, \(a_{1} = 20, a_{0} = 12, a_{2} = 11\) and h = 10
\(Mode = l + \left ( \frac{a_{1}a_{0}}{2a_{1}a_{0}a_{2}} \right )\times h\)
\(= 40 + \left ( \frac{20 – 12}{2 \times 20 – 12 – 11} \right )\times 10\)
\(= 40 + \left ( \frac{8}{17} \right )\times 10 = 44.70\)
Exercise 3
Question 1:
The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption(in units)  No. of customers 
6585  4 
85105  5 
105125  13 
125145  20 
145165  14 
165185  8 
185205  4 
Solution:
Class Interval  Frequency  Cumulative frequency 
6585  4  4 
85105  5  9 
105125  13  22 
125145  20  42 
145165  14  56 
165185  8  64 
185205  4  68 
N=68 
Where, n = 68 and hence \(\frac{n}{2}=34\)
Hence, the median class is 125145 with cumulative frequency = 42
Where, l = 125, n = 68, cf = 22, f = 20, h = 20
Median is calculated as follows:
Median=\(l+\left ( \frac{\frac{n}{2}cf}{f} \right )Xh\)
=\(125+\left ( \frac{3422}{20} \right )X20\)
=125+12=137
Mode = Modal class = 125145, \(f_{1} = 20, f_{0} = 13, f_{2} = 14\) & h = 20
Mode=\(l+\left ( \frac{f_{1}f_{0}}{2Xf_{1}f_{0}f_{2}} \right )Xh\)
=\(125+\left ( \frac{2013}{2X201314} \right )X20\)
=\(125+\frac{7}{13}X20\)
=125+10.77
=135.77
Calculate the Mean:
Class Interval  f_{i}  x_{i}  d_{i}=x_{i}a  u_{i}=d_{i}/h  f_{i}u_{i} 
6585  4  75  60  3  12 
85105  5  95  40  2  10 
105125  13  115  20  1  13 
125145  20  135  0  0  0 
145165  14  155  20  1  14 
165185  8  175  40  2  16 
185205  4  195  60  3  12 
Sum f_{i}= 68  Sum f_{i}u_{i} = 7 
\(\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}X h\)
\(=135+\frac{7}{68}X20\)
=137.05
Mean, median and mode are more/less equal in this distribution.
Question 2:
If the median of a distribution given below is 28.5 then, find the value of an x &y.
Class Interval  Frequency 
010  5 
1020  x 
2030  20 
3040  15 
4050  y 
5060  5 
Total  60 
Solution:
n = 60
Where,\(\frac{n}{2}\) = 30
Median class is 20 – 30 with a cumulative frequency = 25 + x
Lower limit of median class = 20, cf = 5 + x , f = 20 & h = 10
Median=l+\(\left ( \frac{\frac{n}{2}cf}{f} \right )Xh\)
Or,28.5=20+\(\left ( \frac{305x}{20} \right )X10\)
Or,\(\frac{25x}{2}\)
Or, 25x=17
Or, x= 25 – 17 =8
Now, from cumulative frequency, we can identify the value of x + y as follows:
60=5+20+15+5+x+y
Or, 45+x+y=60
Or, x + y=6045=15
Hence, y = 15 – x= 15 – 8 = 7
Hence, x = 8 & y = 7
Question 3:
The Life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the persons whose age is 18 years onwards but less than the 60 years.
Age(in years)  Number of policy holder 
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 
Solution:
Class interval  Frequency  Cumulative frequency 
1520  2  2 
2025  4  6 
2530  18  24 
3035  21  45 
3540  33  78 
4045  11  89 
4550  3  92 
5055  6  98 
5560  2  100 
Where, n = 100 and \(\frac{n}{2}\) = 50
Median class = 3545
Then, l = 35, cf = 45, f = 33 & h = 5
Median=l+\(\left ( \frac{\frac{n}{2}cf}{f} \right )Xh\)
\(=35+\left ( \frac{5045}{33} \right )X5\)
\(=35+\left ( \frac{25}{33} \right )\)
=35.75
Question 4:
The lengths of 40 leaves in a plant are measured correctly to the nearest millimetre, and the data obtained is represented as in the following table:
Length(in mm)  Number of leaves 
118126  3 
127135  5 
136144  9 
145153  12 
154162  5 
163171  4 
172180  2 
Find the median length of leaves.
Solution:
Class Interval  Frequency  Cumulative frequency 
117.5126.5  3  3 
126.5135.5  5  8 
135.5144.5  9  17 
144.5153.5  12  29 
153.5162.5  5  34 
162.5171.5  4  38 
171.5180.5  2  40 
Where, n = 40 and \(\frac{n}{2}\) = 20
Median class = 144.5153.5
then, l = 144.5, cf = 17, f = 12 & h = 9
Median=l+\(\left ( \frac{\frac{n}{2}cf}{f} \right )Xh\)
\(=144.5+\left ( \frac{2017}{12} \right )X9\)
\(=144.5+\left ( \frac{9}{4} \right )\)
=146.75
Question 5:
The following table gives the distribution of a life time of 400 neon lamps.
Lifetime(in hours)  Number of lamps 
15002000  14 
20002500  56 
25003000  60 
30003500  86 
35004000  74 
40004500  62 
45005000  48 
Find the median lifetime of a lamp.
Solution:
Class Interval  Frequency  Cumulative 
15002000  14  14 
20002500  56  70 
25003000  60  130 
30003500  86  216 
35004000  74  290 
40004500  62  352 
45005000  48  400 
Where, n = 400 &\(\frac{n}{2}\) = 200
Median class = 3000 – 3500
Therefore, l = 3000, cf = 130, f = 86 & h = 500
Median=l+\(\left ( \frac{\frac{n}{2}cf}{f} \right )Xh\)
\(=3000+\left ( \frac{200130}{86} \right )X500\)
=3000+406.97
=3406.97
Question 6:
In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:
Number of letters  14  47  710  1013  1316  1619 
Number of surnames  6  30  40  16  4  4 
Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.
Solution: To Calculate median:
Class Interval  Frequency  Cumulative Frequency 
14  6  6 
47  30  36 
710  40  76 
1013  16  92 
1316  4  96 
1619  4  100 
Where, n = 100 &\(\frac{n}{2}\) = 50
Median class = 710
Therefore, l = 7, cf = 36, f = 40 & h = 3
Median=l+\(\left ( \frac{\frac{n}{2}cf}{f} \right )Xh\)
\(=7+\left ( \frac{5036}{40} \right )X3\)
\(=7+\left ( \frac{14}{40} \right )X3\)
=8.05
Calculate the Mode:
Modal class = 710,
Where, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3
Mode=\(l+\left ( \frac{f_{1}f_{0}}{2Xf_{1}f_{0}f_{2}} \right )Xh\)
=\(7+\left ( \frac{4030}{2X403016} \right )X3\)
=\(7+\frac{10}{34}X3\)
=7.88
Calculate the Mean:
Class Interval  f_{i}  x_{i}  f_{i}x_{i} 
14  6  2.5  15 
47  30  5.5  165 
710  40  8.5  340 
1013  16  11.5  184 
1316  4  14.5  51 
1619  4  17.5  70 
Sum f_{i} = 100  Sum f_{i}x_{i} = 825 
\(\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}X h\)
\(=\frac{825}{100}\)
=8.25
Q7. The distributions of below gives a weight of 30 students of a class. Find the median weight of a students.
Weight(in kg)  4045  4550  5055  5560  6065  6570  7075 
Number of students  2  3  8  6  6  3  2 
Solution:
Class Interval  Frequency  Cumulative frequency 
4045  2  2 
4550  3  5 
5055  8  13 
5560  6  19 
6065  6  25 
6570  3  28 
7075  2  30 
Where, n = 30 and \(\frac{n}{2}\)= 15
median class = 5560
therefore, l = 55, cf = 13, f = 6 & h = 5
Median=l+\(\left ( \frac{\frac{n}{2}cf}{f} \right )Xh\)
\(=55+\left ( \frac{1513}{6} \right )X5\)
\(=55+\left ( \frac{5}{3} \right )\)
=56.67
Question 7:
The distribution shows the pocket money of students in an area. The mean pocket money is Rs 18. Find the missing frequency f.
Daily pocket allowance (in Rs)  1113  1315  1517  1719  1921  2123  2325 
Number of children  07  06  09  013  F  05  04 
Solution:
Class interval  f_{i}  x_{i}  f_{i}x_{i} 
1113  07  012  84 
1315  06  014  84 
1517  09  016  144 
1719  013  018  234 
1921  f  020  20f 
2123  05  022  110 
2325  04  024  96 
Sum f_{i} = 44+f  Sum f_{i}x_{i} = 752+20f 
We have,
Mean pocket money, \(x_{i}\) = 18(given)
Hence substituting for f_{i}x_{i},
18(44+f) = 752 + 20f
Or, 792 + 18f = 752 + 20f
Or, 2f = 40
\( \rightarrow f=20\)
The modern world is highly dataoriented and it becomes necessary to know to represent data in a meaningful way. The branch of mathematics which deals with the representation of data in a meaningful way is known as statistics. Let us look at a few examples of how statistics is useful in our daily life
 The data that you see in the weather forecast section of the news is due to the goodwill of statistics.
 The selection of a player in any sport is mostly made based on the player’s statistical performance report.
In this chapter, students are introduced to histograms, graphical representations of data in the form of bar charts, cumulative frequency tables and frequency polygons. Students will be introduced to the three measures of central tendency. The three measures of central tendency are mean, median and mode.
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