Notes
Assignments
NCERT Solutions
Flash back from IX Class
A systematic record of facts or different values of a quantity is called
data.
Features of the data
 Statistics deals with collection, presentation, analysis and interpretation of numerical data.
 Arranging data in a order to study their salient features is called presentation of data.
 Data arranged in ascending or descending order is called arrayed data or an array
 Range of the data is the difference between the maximum and the minimum values of the observations
 Table that shows the frequency of different values in the given data is called a frequency distribution table
 A frequency distribution table that shows the frequency of each individual value in the given data is called an ungrouped frequency distribution table.
 A table that shows the frequency of groups of values in the given data is called a grouped frequency distribution table
 The groupings used to group the values in given data are called classes or classintervals. The number of values that each class contains is called the class size or class width. The lower value in a class is called the lower class limit. The higher value in a class is called the upper class limit.
 Class mark of a class is the mid value of the two limits of that class.
 A frequency distribution in which the upper limit of one class differs from the lower limit of the succeeding class is called an Inclusive or discontinuous Frequency Distribution.
 A frequency distribution in which the upper limit of one class coincides from the lower limit of the succeeding class is called an exclusive or continuous Frequency Distribution
Bar Graph:
A bar graph is a pictorial representation of data in which rectangular bars of uniform width are drawn with equal spacing between them on one axis, usually the x axis. The value of the variable is shown on the other axis that is the y axis.
Histogram:
A histogram is a set of adjacent rectangles whose areas are proportional to the frequencies of a given continuous frequency distribution
Mean
The mean value of a variable is defined as the sum of all the values of the variable divided by the number of values.
Median
The
median of a set of data values is the middle value of the data set when it has been arranged in ascending order. That is, from the smallest value to the highest value
Median is calculated as
Where n is the number of values in the data. If the number of values in the data set is even, then the
median is the average of the two middle value
Mode
Mode of a statistical data is the value of that variable which has the maximum frequency
Mean for Ungroup Frequency table
Here is the ungroup Frequency table
Mark obtained(x_{i})

25

35

45

65

No of student(f_{i)}

4

10

23

34

ean is given by
Greek letter
∑ (capital sigma) means summation
Example
A survey was conducted by a group of students as a part of their environment awareness programmes, in which they collected the following data regarding the number of plants in 30 houses in a locality. Find the mean number of plants per house.
Number of Plants

1

3

5

7

9

11

13

Number of Houses

11

2

1

5

6

2

3

Solution
Number of Plants

No. of houses (f_{i})

f_{i} x_{i}

1

11

11

3

2

6

5

1

5

7

5

35

9

6

54

11

2

22

13

3

39


∑f_{i }= 30

∑f_{i} x_{i} = 172

ean = 172/30 = 5.73
Mean for group Frequency table
Class interval

1025

2545

4565

6585

No of student(f_{i)}

4

10

23

34

In these distribution, it is assumed that frequency of each class interval is centered around its midpoint i.e class marks
Mean can be calculated using three method
a) Direct method
This method can be very calculation intensive if the values of f and x are large.We have big calculations and chance of making mistake is quite high
Steps involved in finding the mean using Direct Method
1) Prepare a frequency table with the help of class marks
2) Multiply f
_{i } x
_{i } and find the sum of it.
4) Use the above formula and find the mean.
Example
The following table shows the weights of 10 children:
Weight (in kg)

6668

6870

7072

7274

7476

Number of students

3

3

2

1

1

Find the mean by using direct method.
Solution:
Weight (in kg)

x_{i}

No. of students (f_{i})

f_{i} x_{i}

6668

67

3

201

6870

69

3

207

7072

71

2

142

7274

73

1

73

7476

75

1

75



Σ f_{i} = 10

Σf_{i} x_{i} = 698

So, Mean would be
=698/10 = 69.8 kg
b) Assumed mean method
Where
a=> Assumed mean
d
_{i } => x
_{i} –a
This method is quite useful when the values of f and x are large. It makes the calculation easiar.In this method we take some assumed mean and calculate the deviation from it and then calculate mean using above formula
Steps involved in finding the mean using Assumed Mean Method
1) Prepare a frequency table.
2) Choose A and take deviations d
_{i } = x
_{i }  A of the values of x
_{i }.
3) Multiply f
_{i } d
_{i } and find the sum of it.
4) Use the above formula and find the mean.
Example
Example
The following table shows the weights of 10 children:
Weight (in kg)

6668

6870

7072

7274

7476

Number of students

3

3

2

1

1

Find the mean by using Assumed Mean method.
Solution:
Let the assumed mean = A = 71
Weight (in kg)

x_{i}

No. of students (f_{i})

di = xi  71

f_{i} d_{i}

6668

67

3

4

 12

6870

69

3

2

 6

7072

71

2

0

0

7274

73

1

2

2

7476

75

1

4

4



Σ fi = 10


Σfi di = 12

So, Mean would be
=7112/10 = 69.8 kg
c) Step deviation Method
Where
a=> Assumed mean
u
_{i } => (x
_{i} –a)/h
This method is quite useful when the values of f and x are large. It makes the calculation further easiar by dividing the deviation from common factor.
Steps involved in finding the mean using Step Deviation Method
1) Prepare a frequency table.
2) Choose A and h and take u
_{i } = (x
_{i} –a)/h of the values of x
_{i }.
3) Multiply f
_{i } u
_{i } and find the sum of it.
4) Use the above formula and find the mean.
Example
The following table shows the weights of 10 children:
Weight (in kg)

6668

6870

7072

7274

7476

Number of students

3

3

2

1

1

Find the mean by using Step Deviation method.
Solution:
Let the assumed mean = A = 71 and h=2
Weight (in kg)

x_{i}

No. of students (f_{i})

di = xi  71

u_{i} =d_{i}/h

f_{i} u_{i}

6668

67

3

4

2

 6

6870

69

3

2

1

 3

7072

71

2

0

0

0

7274

73

1

2

1

1

7476

75

1

4

2

2



Σ fi = 10



Σfi ui = 6

So, Mean would be
=71+ (6/10) 2 = 69.8 kg
Important points
1)The mean obtained by all these three methods are same.
2) The assumed mean method and stepdeviation method are just simplified forms of the direct method.
Mode for grouped frequency table
Modal class: The class interval having highest frequency is called the modal class and Mode is obtained using the modal class
Where
l = lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
f
_{1} = frequency of the modal class,
f
_{0} = frequency of the class preceding the modal class,
f
_{2} = frequency of the class succeeding the modal class.
Example
The following table shows the ages of the patients admitted in a hospital during a year
Age (in years)

515

1525

2535

3545

4555

5565

Number of patients

6

11

21

23

14

5

Find the mode
Solution
odal class = 35 – 45, l = 35, class width (h) = 10, f
_{1} = 23, f
_{0} = 21 and f
_{2} = 14
Substituting the values in the formula given above we get
ode= 36.8 year
Cumulative Frequency chart
The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.
Class interval ( Age)

No of Insurance policies

1520

2

2025

4

2530

16

3035

20

3540

20

4045

12

Cumulative Frequency chart will be like
Age in years

Cumulative No of Insurance policies

Less than 20 years

2

Less than 25 years

6

Less than 30 years

22

Less than 35 years

42

Less than 40 years

62

Less than 45 years

74

The above table cumulative frequency distribution of the less than type. We can similary make it like below
Age in years

Cumulative No of Insurance policies

More than or equal to 15 years

74

More than or equal to 20 years

742=72

More than or equal to 25 years

724=68

More than or equal to 30 years

6816=52

More than or equal to 35 years

5220=32

More than or equal to 40 years

3220=12

The table above is called a cumulative frequency distribution of the more than type.
Steps involved in finding Median of a grouped data frequency table
1) For the given data, we need to have class interval, frequency distribution and cumulative frequency distribution
2)Then we need to find the median class
How to find the median class
a) we find the cumulative frequencies of all the classes and n/2
b)We now locate the class whose cumulative frequency is greater than (and nearest to) n/2
c)That class is called the median class
3) Median is calculated as
Where
l = lower limit of median class,
n = number of observations,
cf = cumulative frequency of class preceding the median class,
f = frequency of median class,
h = class size (assuming class size to be equal)
Example
A survey regarding the heights (in cm) of 60 girls of a school was conducted and the following data was obtained:
Height (in cm)

Number of girls

Less than 140

4

Less than 145

11

Less than 150

29

Less than 155

40

Less than 160

46

Less than 165

60

Find the median height
Solution
To calculate the median height, we need to find the class intervals and their corresponding frequencies.
The given distribution being of the less than type, 140, 145, 150, . . ., 165 given the upper limits of the corresponding class intervals.
So, the classes should be below 140, 140  145, 145  150, . . ., 160  165. Observe that from the given distribution, we find that there are 4 girls with height less than 140, i.e., the frequency of class interval below 140 is 4 . Now, there are 11 girls with heights less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval 140145 will be 114=7. Similarly, other can be calculated
Class interval

Frequency

Cumulative Frequency

Below 140

4

4

140145

7

11

145150

18

29

150155

11

40

155 160

6

46

160 165

14

60

So, n =60 and n/2=30 And cumulative frequency which is greater than and nearest to 30 is 40 , So median class 150155
l (the lower limit) = 150,
cf (the cumulative frequency of the class preceding 150  155) = 29,
f (the frequency of the median class 150  151) = 11,
h (the class size) = 5.
Now
= 150 + [(3029)/11]5
=150.45 cm
3 Median=Mode +2 Mean
Graphical representation of Cummulative frequency distribution
We can represent Cummulative frequency distribution on the graph also. To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (xaxis) and their corresponding cumulative frequencies on the vertical axis (yaxis), choosing a convenient scale.
When we draw the graph for the cumulative frequency distribution of the less than type.The curve we get is called a cumulative frequency curve, or an ogive (of the less than type).
When we draw the graph for the cumulative frequency distribution of the more than type.The curve we get is called a cumulative frequency curve, or an ogive (of the more than type).
When we plot both these curve on the same axis, The two ogives willintersect each other at a point. From this point, if we draw a perpendicular on the xaxis, the point at which it cuts the xaxis gives us the median
Crossword Puzzle
Across
3. The ........ frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.
5. It is the difference between the maximum and minimum values in data set.
7. The highest frequency class interval is .....class
8. Middle value of the dataset
Down
1. Terms for number of times the event occurred in an experiment or study
2. It is a set of adjacent rectangles whose areas are proportional to the frequencies of a given continuous frequency distribution
4. Maximum frequency data in data set
6. Average value of a data set
Check your Answers
1)Frequncy
2)Histogram
3)Cumulative
4)Mode
5)Range
6)Mean
7)Model
8)median
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Class 10 Maths
Class 10 Science