Statistics: Mean, Median, Variance and Cumulative Frequency Curve

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Jasmine Grover

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Statistics is a branch of mathematics that is focused on collecting, analyzing, organizing, and interpreting different types of data. It holds immense value in scientific, economic, social, and industrial studies.


Types of Data

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  • Ungrouped Data:

The data which is present in its original or unsorted form is called ungrouped data. It is the set of data that you initially get from a survey or experiment.

Example: Let’s say the marks of 15 students present in a class are:

18, 14, 8, 10, 18, 19, 16, 15, 18, 7, 15, 19, 10, 17, 12

From this data, we can interpret that 19 is the highest mark and two students have got the highest marks. It is easy to work on ungrouped data if the number of data is small.

  • Grouped Data:

The data which is sorted into different categories or groups is called grouped data.

For example, we have the marks obtained by 50 students in an examination. Such huge data is difficult to interpret in an ungrouped manner. So we sort the data as given in the table below:

Marks Obtained

1-10

10-20

20-30

30-40

40-50

No. Of Students

5

10

16

14 

5

  • Frequency

The total number of times that a specific observation is present in data is called frequency. 

  • Class interval

Class interval is used to group the observations into specific classes. Considering the table given above, we have grouped the number of students into the marks interval. Here, the marks interval is the class interval. 

Class width= upper limit of class - the lower limit of class

Mean, Median and Mode Video Explanation


Measures Of Central Tendency

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Mean

In simple terms, the average of a data set is called the mean.

For ungrouped data, 

Mean = sum of observations÷ total number of observations

Mean= 

x=x1 + + x2 + x3+ x4+……+ xnn

For grouped data, there are three methods of finding the mean. 

  • Direct method:

X=∑fixi/fi

Where fi= frequency of ith class with a class mark of xi.

Class mark: Class mark is the midpoint of a class that contains the frequency.

Class mark =(upper limit of class + lower limit of class)/2

  • Assume mean method:

In this method, we have to select one of the numbers generally present in the middle, as a supposed mean, 'a'.

We have to find the value of 'd' that is the deviation of 'a' from all the observations xi. 

  • Step deviation method:

Find the class mark of the class interval by taking a point of lower and upper-class limits. (Formula of classmark is mentioned ready)

Now take any value of xi, preferably from the middle, and represent it with “a”.

With the formula given below, find the deviation of a:

di=xi-a

Now you have to find out the mean of the deviations with the formula given below:

d? =∑xidi/∑fi

Finally, calculate the mean:

x? = a + ∑xidi/∑fi

Median

The middlemost observation of a set of data is called the median. Methods of finding median are:

  • Median of grouped data without class intervals:
  • Arrange all the observations and frequencies in ascending order. 
  • Add all the frequencies up to a certain observation to find the cumulative frequency of that observation. 
  • For an odd number of observations, the median is the observation whose cumulative frequency is just larger than or equal to (n+1)/2.
  • The median is the average of observations whose cumulative frequency is just larger than or equal to n/2 and (n/2)+1, in the case of an even number of observations.
  • Median of grouped data with class interval:

It is not possible to find the median of grouped data by observing the frequency table, so we have to use the formula given below:

Median= l + (n/2-cf )/f X h

Where,

n= number of observations,

l= the lower limit of median class,

cf=the cumulative frequency of the class that precedes the median class,

h= class width,

f= frequency of median class


Graphical method of finding median:

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To find the median through the graph, first, you have to draw the graph properly. Follow the steps below:

1.The first step of finding the median is to determine the median class.

2.Label the y-axis with cumulative frequencies and the x-axis with data belonging to the median class.

3.Connect the extremes of class and cumulative frequencies with a straight line graph.

4.Locate the graph point that corresponds to cf=n/2.

5.From this point on t, drop a perpendicular.

A cumulative frequency curve, also known as an ogive, is created by linking the points.

Ogives are divided into two categories:

  • Less Than Ogive: Cumulative frequencies of 'less than' are displayed against upper-class limits of distinct class intervals. It is usually an ascending curve.
  •  More Than Ogive: The cumulative frequencies of 'more than' are displayed against the lower class limits. It is usually a decreasing curve.

Find the point at which both ogive types intersect and this point denotes the median of the grouped data.

Mode

The observation or number which occurs most often or frequently in a data set is called mode. 

  •  Mode of ungrouped data:

Finding the mode of ungrouped data is very easy. Just notice which observation has the maximum frequency and that observation is considered to be the mode of that particular set of data.

For example, consider the numbers:

7, 8, 18, 7, 3, 20, 3, 18, 7

The number 7 occurs most frequently, therefore, mode=7.

  • Mode of Grouped Data:

You can find the mode of grouped data using the following formula:

l=(f1-f0/2f1-f0-f2)X h

where l= lower limit of modal class,

h= class width,

f0= frequency of the class that precedes the median class

f1= frequency of the modal class,

f2= frequency of the class that succeeds the modal class.


Previous Years Questions

Ques: The median of the data, in a continuous frequency distribution is given as 21. If each and every observation of the data is increased by a value of 5, then find the new median. (CBSE 2015)

Answer

Since each data is increased by an equal amount, i.e, 5, the median will also be increased by 5.

Therefore, new median = 21 + 5 = 26

Ques: Following table shows the sale of shoes in a store during one month:

Size of shoe

No. of pairs sold

3

4

4

18

5

25

6

12

7

5

8

1

Find the model size of the shoes sold. (CBSE 2014)

Answer.

The modal size of the shoes would be the modal class and the model class is the class with the highest frequency.

Maximum no. of pairs sold = 25 which is of size 5.

∴ Modal size of shoes = 5

Ques: The mean of the following data given in a table is 18.75. Find the value of P. (CBSE 2012, 2017)

Class marks (xi

Frequency(fi

10

5

5

10

P

7

25

8

30

2

Answer.

Class marks xi 

Frequency fi

fixi

10

5

50

5

10

50

P

7

7P

25

8

200

30

2

60

∑fi = 32

∑fixi = 360+7P

We know, Mean = ∑fixi/∑fi 

Put the values of mean=18.75 (given), ∑fixi, and∑fi 

18.75/1 = 360+7P/32

360 + 7P=600

7P=600-360 = 240

P=240/7 = 34.29 (Ans)

Ques: When it is given that mode = 35.3 and mean = 30.5, using an empirical formula, find the median of the data. (CBSE 2014)

Answer.

We know that, empirical formula: Mode = 3(Median) – 2Median

Therefore, 35.3 = 3(Median) – 2(30.5)

=> 35.3 = 3(Median) – 61

=> 96.3 = 3 Median

=> Median = 96.33/3 = 32.1

Ques: If the median of the given frequency distribution is 32.5. Find the values of f1 and f2. (CBSE 2019)

Class

0-10

10-20

20-30

30-40

40-50

50-60

60-70

Total

Frequency

f

5

9

12

f2

3

2

40

Answer.

Class

Frequency

Cumulative Frequency (C.F)

0-10

f1

f1

10-20

5

5+f1

20-30

9

14+f1

30-40

12

26+f1

40-50

f2

26+f1 + f2

50-60

3

29+f1 + f2

60-70

2

31+f1 + f2

Total:

40

Since median = 32.5, median class is 30-40.

32.5 = 30 + 10/12 (20-14-f1)

=>f1=3

Also, 31+f1+f2=40

=> 34+f2=40

=> f2= 6

CBSE X Related Questions

1.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

      2.
      Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
      (i) 2, 4, 8, 16, . . . .
      (ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
      (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
      (iv) – 10, – 6, – 2, 2, . . .
      (v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
      (vi) 0.2, 0.22, 0.222, 0.2222, . . . .
      (vii) 0, – 4, – 8, –12, . . . .
      (viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
      (ix) 1, 3, 9, 27, . . . .
      (x) a, 2a, 3a, 4a, . . . .
      (xi) a, \(a^2, a^3, a^4,\)  . . . .
      (xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
      (xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
      (xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
      (xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

          3.
          Find the sums given below :
          1. \(7 + 10\frac 12+ 14 + ....... + 84\)
          2. \(34 + 32 + 30 + ....... + 10\)
          3. \(–5 + (–8) + (–11) + ....... + (–230)\)

              4.

              Form the pair of linear equations for the following problems and find their solution by substitution method.

              (i) The difference between two numbers is 26 and one number is three times the other. Find them.

              (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

              (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

              (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.

              (v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

              (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

                  5.

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

                  Length (in mm)

                  Number of leaves

                  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.)

                      6.
                      The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

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