Statistics: Mean, Median, Ogive, Mode

The arithmetic mean is important in statistics. It is also called average or average value, is the quantity obtained by summing two or more numbers or variables and then dividing with the number of numbers or variables.

Direct method of finding the arithmetic mean is the easiest.

We have been given frequency distribution table, let’s make it again.

By direct method, arithmetic mean (AM) is given by

⇒

⇒ AM = 17.425

Thus, average age of his/her friends is 17.4.

Since arithmetic mean is also called average or average value, we need to find the average.

So, we need to create a table showing the frequency.

Observe that, we have taken assumed mean as 4. Assumed mean is a mid-value usually from the x_i’s.

Here, d_i is the deviation from the mean.

Using x_i’s and A, we have found d_i and further have found out f_id_i.

We have the following formula to calculate arithmetic mean using assumed-mean method,

Substituting all the values from the table into the formula, we get

⇒

⇒

⇒ Mean = 4.24

Thus, the average members of 50 families is 4.24.

We have been given the arithmetic mean of the data, that is, 20.6.

⇒ Mean = 20.6

To find the value of a, we need to simply find mean using direct method.

So, let’s make a table showing frequencies and f_ix_i.

So, we get

∑f_ix_i = 530 + 25a

∑f_i = 50

Using direct method, we have the formula of arithmetic mean as follows:

⇒ [∵ Mean = 20.6, ∑f_ix_i = 530 + 25a & ∑f_i = 50]

⇒ 1030 = 530 + 25a

⇒ 25a = 1030 – 530

⇒ 25a = 500

⇒

⇒ a = 20

Thus, a = 20.

We have been given the arithmetic mean of the data, that is, 15.

⇒ Mean = 15

To find the value of p, we need to simply find mean using direct method.

So, let’s make a table showing frequencies and f_ix_i.

So, we get

∑f_ix_i = 445 + 10p

∑f_i = 27 + p

Using direct method, we have the formula of arithmetic mean as follows:

⇒ [∵ Mean = 15, ∑f_ix_i = 445 + 10p & ∑f_i = 27 + p]

⇒ 15 (27 + p) = 445 + 10p

⇒ 405 + 15p = 445 + 10p

⇒ 15p – 10p = 445 – 405

⇒ 5p = 40

⇒

⇒ p = 8

Thus, p = 8.

Arithmetic mean of grouped frequency distribution is found by taking out midpoints of the class intervals.

Let’s create a table showing frequencies and f_ix_i.

So, we have

∑f_ix_i = 2736

∑f_i = 50

Using the direct method, we have the formula for arithmetic mean.

Using the above values, we get

⇒ Mean = 54.72

Thus, mean is 54.72.

We need to find the average using any method of calculation.

For the data given, the average can be calculated using direct method as the data is small. Using direct method for the given answer will provide us easier and faster calculation.

Let’s create a table showing frequencies and f_ix_i.

So, we have

∑f_ix_i = 4340

∑f_i = 100

Using the direct method, we have the formula for arithmetic mean.

Using the above values, we get

⇒ Mean = 43.4

Thus, average is 43.4.

Arithmetic mean is just the average value of the data or variables given in the data.

To find the mean using direct method, let’s create a table showing frequencies and f_ix_i.

So, we have

∑f_ix_i = 750

∑f_i = 30

Using the direct method, we have the formula for arithmetic mean.

Using the above values, we get

⇒ Mean = 25

Thus, mean is 25.

Arithmetic mean is just the average value of the data or variables given in the data.

Let’s create a table showing frequencies and f_ix_i.

So, we have

∑f_ix_i = 4030

∑f_i = 100

Using the direct method, we have the formula for arithmetic mean.

Using the above values, we get

⇒Mean = 40.3

Thus, mean is 40.3.

We need to find the mean by assumed mean method.

For assumed-mean method, we assume a mean from the midpoints and calculate deviation from that assumed mean.

We can create a table showing frequencies, midpoints and deviations.

So, we have

∑f_ix_i = 9080

∑f_i = 90

Observe that, we have taken assumed mean as 100. Assumed mean is a mid-value usually from the x_i’s.

Here, d_i is the deviation from the mean.

Using x_i’s and A, we have found d_i and further have found out f_id_i.

We have the following formula to calculate arithmetic mean using assumed-mean method,

Substituting all the values from the table into the formula, we get

⇒

⇒

⇒ Mean = 200.89

Thus, the mean of the data is 200.89.

We need to find the mean by assumed mean method.

For assumed-mean method, we assume a mean from the midpoints and calculate deviation from that assumed mean.

We can create a table showing frequencies, midpoints and deviations.

So, we have

∑f_ix_i = 2060

∑f_i = 40

Observe that, we have taken assumed mean as 50. Assumed mean is a mid-value usually from the x_i’s.

Here, d_i is the deviation from the mean.

Using x_i’s and A, we have found d_i and further have found out f_id_i.

We have the following formula to calculate arithmetic mean using assumed-mean method,

Substituting all the values from the table into the formula, we get

⇒

⇒

⇒ Mean = 101.5

Thus, the mean of the data is 101.5.

To find mean using step-deviation method, we need to create a table showing frequencies, midpoints and deviations.

Observe that class size, h = (30 – 0) = (60 – 30) = … = (150 – 120) = 30

We have,

A = 75

∑f_iu_i = 2

∑f_i = 80

h = 30

By step deviation method, we can give the formula of mean as:

Using the above values, we have

⇒

⇒

⇒

⇒

⇒Mean = 75.75

Thus, mean is 75.75.

To find mean using step-deviation method, we need to create a table showing frequencies, midpoints and deviations.

Observe that class size, h = (14 – 0) = (28 – 14) = … = (70 – 56) = 14

We have,

A = 35

∑f_iu_i = 8

∑f_i = 90

h = 14

By step deviation method, we can give the formula of mean as:

Using the above values, we have

⇒

⇒

⇒

⇒ Mean = 36.24

Thus, mean is 36.24.

Given that, mean = 24

We need to find the value of p.

To find p, we need to calculate mean by direct method.

Let us make a table showing frequencies, midpoints and f_ix_i.

We have

∑f_ix_i = 1700 + 35p

∑f_i = 80 + p

By direct method, we have the formula of arithmetic mean:

⇒

⇒ 24 × (80 + p) = 1700 + 35p

⇒ 1920 + 24p = 1700 + 35p

⇒ 35p – 24p = 1920 – 1700

⇒ 11p = 220

⇒

⇒ p = 20

Thus, p = 20.

The arithmetic mean is also called average or average value, is the sum of the data or variables divided by total number of data or variables.

This is an inclusive type of data, so in order to find the average or arithmetic mean we need to convert it into exclusive type of data.

To convert inclusive type of data into exclusive type, subtract 0.5 from the lower class limit and add 0.5 to upper class limit.

For example, in class limit 30 – 34:

Subtract 0.5 from lower class limit and add 0.5 to upper class limit.

30 – 0.5 = 29.5

34 + 0.5 = 34.5

So, the class interval comes out to be 29.5 – 34.5.

Similarly, we have with other class intervals.

Let’s make a table showing exclusive type of data.

So, we have

∑f_ix_i = 2055

∑f_i = 50

From direct method, mean is given by

Using the above values,

⇒ Mean = 41.1

Thus, mean is 41.1.

The arithmetic mean is also called average or average value, is the sum of the data or variables divided by total number of data or variables.

This is an inclusive type of data, so in order to find the average or arithmetic mean we need to convert it into exclusive type of data.

To convert inclusive type of data into exclusive type, subtract 0.5 from the lower class limit and add 0.5 to upper class limit.

For example, in class limit 5 – 14:

Subtract 0.5 from lower class limit and add 0.5 to upper class limit.

5 – 0.5 = 4.5

14 + 0.5 = 14.5

So, the class interval comes out to be 4.5 – 14.5.

Similarly, we have with other class intervals.

Let’s make a table showing exclusive type of data.

So, we have

∑f_ix_i = 2140

∑f_i = 60

From direct method, mean is given by

Using the above values,

⇒ Mean = 35.67

Thus, mean is 35.67.

We have been given cumulative frequencies of data as given in question.

We need to convert it into grouped frequency distribution (exclusive type of data).

So, let us create a table showing the frequencies and class intervals.

So, we have

∑f_ix_i = 3625

∑f_i = 105

From direct method, mean is given by

Using the above values,

⇒ Mean = 34.52

Thus, mean is 34.52.

Observe, here the class intervals are unequal but continuous. So, in order to find mean we need to simply find midpoints of the class intervals and further find f_ix_i.

We can solve this using three methods,

- Direct method

- Assumed-mean method

- Step Deviation method

We shall calculate the mean using direct method.

So, we have

∑f_ix_i = 748

∑f_i = 64

From direct method, mean is given by

Using the above values,

⇒ Mean = 11.69

Thus, mean is 11.69.

To find median, it is necessary to arrange the data in either ascending or descending order.

So, let’s arrange the data into ascending order. It is,

52, 75, 92, 107, 113, 195, 201

Note that, total number of data, n = 7

Here, n is odd.

When n is odd, median is found out as

Putting n = 7, we get

⇒

⇒ Median = 4^th term

Note in the ascended order list, 4^th term is 107.

∵ 1^st term = 52,

2^nd term = 75,

3^rd term = 92,

4^th term = 107 ß

Thus, median is 107.

To find median, it is necessary to arrange the data in either ascending or descending order.

So, let’s arrange the data into ascending order. It is,

4, 5, 6, 9, 10, 11, 18, 20

Note that, total number of data, n = 8

Here, n is even. When n is even, median is found out as

Putting n = 8, we get

⇒

⇒

⇒

⇒ Median = 9.5

Thus, median is 9.5.

To find median, it is necessary to arrange the data in either ascending or descending order.

So, let’s arrange the data into ascending order. It is,

42, 45, 45, 50, 51, 52, 54, 55, 56, 58, 59, 60, 62, 64

Note that, total number of data, n = 14

Here, n is even.

When n is even, median is found out as

Putting n = 14, we get

⇒

⇒

⇒

⇒ Median = 54.5

Thus, median is 54.5.

To find median, it is necessary to arrange the data in either ascending or descending order.

So, let’s arrange the data into ascending order. It is,

6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 11, 11

Note that, total number of data, n = 22

Here, n is even.

When n is even, median is found out as

Putting n = 22, we get

⇒

⇒

⇒

⇒ Median = 8

Thus, median is 8.

To find out median, we need to calculate cumulative frequency for the given data.

Let us make a table for that.

Here, n = 70, that is, n = even.

Since, n is even, median can be found out as:

⇒

⇒ Median = The mean of 35^th and 36^th observations

And we know that, 35^th observation = 47 [∵ 35 comes between 32 and 44]

And 36^th observation = 47 [∵ 36 comes between 32 and 44]

Median = Mean of 47 and 47

⇒

⇒

⇒ Median = 47

Thus, median = 47.

To find out median, we need to calculate cumulative frequency for the given data.

Let us make a table for that.

Here, n = 80, that is, n = even.

Since, n is even, median can be found out as:

⇒

⇒ Median = The mean of 40^th and 41^th observations

And we know that, 40^th observation = 22 [∵ 40 comes between 33 and 57]

And 41^th observation = 22 [∵ 41 comes between 33 and 57]

Median = Mean of 22 and 22

⇒

⇒

⇒ Median = 22

Thus, median = 22.

To find out median, we need to calculate cumulative frequency for the given data.

Let us make a table for that.

Here, n = 116, that is, n = even.

Since, n is even, median can be found out as:

⇒

⇒ Median = The mean of 58^th and 59^th observations

And we know that, 58^th observation = 2 [∵ 58 comes between 51 and 86]

And 59^th observation = 2 [∵ 41 comes between 51 and 86]

Median = Mean of 2 and 2

⇒

⇒

⇒ Median = 2

Thus, median = 2.

To find median of tiffin allowance, we need to make a table calculating cumulative frequency.

For median:

We have, total frequency, N = 40

N/2 = 40/2 = 20

Observe, cf = 23 is just greater than 20.

Thus, median class = 50 – 55

Median is given by

Where,

L = Lower class limit of median class = 50

N/2 = 20

cf = cumulative frequency of the class preceding median class = 14

f = frequency of the median class = 9

h = class interval of the median class = 5

Substituting these values in the formula of median, we get

⇒

⇒ Median = 50 + 3.33

⇒ Median = 53.33

Thus, the median of tiffin allowance is Rs. 53.33.

To find median of heights of students, we need to make a table calculating cumulative frequency.

For median:

We have, total frequency, N = 100

N/2 = 100/2 = 50

Observe, cf = 57 is just greater than 50.

Thus, median class = 150 – 155

Median is given by

Where,

L = Lower class limit of median class = 150

N/2 = 50

cf = cumulative frequency of the class preceding median class = 35

f = frequency of the median class = 22

h = class interval of the median class = 5

Substituting these values in the formula of median, we get

⇒

⇒ Median = 150 + 3.409

⇒ Median = 153.409

Thus, the median of heights of students is 153.409 cm.

To find median of data, we need to make a table calculating cumulative frequency.

For median:

We have, total frequency, N = 59

N/2 = 59/2 = 29.5

Observe, cf = 36 is just greater than 29.5.

Thus, median class = 30 – 40

Median is given by

Where,

L = Lower class limit of median class = 30

N/2 = 29.5

cf = cumulative frequency of the class preceding median class = 21

f = frequency of the median class = 15

h = class interval of the median class = 10

Substituting these values in the formula of median, we get

⇒

⇒ Median = 30 + 5.67

⇒ Median = 35.67

Thus, the median of data is 35.67.

To find median of data, we need to make a table calculating cumulative frequency.

For median:

We have, total frequency, N = 50

N/2 = 50/2 = 25

Observe, cf = 26 is just greater than 25.

Thus, median class = 15 – 20

Median is given by

Where,

L = Lower class limit of median class = 15

N/2 = 25

cf = cumulative frequency of the class preceding median class = 11

f = frequency of the median class = 15

h = class interval of the median class = 5

Substituting these values in the formula of median, we get

⇒

⇒ Median = 15 + 9.33

⇒ Median = 24.33

Thus, the median of data is 24.33.

The data given in the question is inclusive type of data, we need to convert it into exclusive type of data.

In order to convert it to exclusive type of data, subtract 0.5 from the lower limit of class interval and add 0.5

To find median of data, we need to make a table calculating cumulative frequency.

For median:

We have, total frequency, N = 30

N/2 = 30/2 = 15

Observe, cf = 18 is just greater than 15.

Thus, median class = 15.5 – 20.5

Median is given by

Where,

L = Lower class limit of median class = 15.5

N/2 = 15

cf = cumulative frequency of the class preceding median class = 11

f = frequency of the median class = 7

h = class interval of the median class = 5

Substituting these values in the formula of median, we get

⇒

⇒ Median = 15.5 + 2.857

⇒ Median = 18.357

Thus, the median of data is 18.357.

The data given in the question is inclusive type of data, we need to convert it into exclusive type of data.

In order to convert it to exclusive type of data, subtract 0.5 from the lower limit of class interval and add 0.5

To find median of data, we need to make a table calculating cumulative frequency.

For median:

We have, total frequency, N = 68

N/2 = 68/2 = 34

Observe, cf = 49 is just greater than 34.

Thus, median class = 80.5 – 90.5

Median is given by

Where,

L = Lower class limit of median class = 80.5

N/2 = 34

cf = cumulative frequency of the class preceding median class = 29

f = frequency of the median class = 20

h = class interval of the median class = 10

Substituting these values in the formula of median, we get

⇒

⇒ Median = 80.5 + 2.5

⇒ Median = 83

Thus, the median of data is 83.

The data given in the question is exclusive type of data.

To find median of data, we need to make a table calculating cumulative frequency.

For median:

We have, total frequency, N = 622

N/2 = 622/2 = 311

Observe, cf = 391 is just greater than 311.

Thus, median class = 60 – 70

Median is given by

Where,

L = Lower class limit of median class = 60

N/2 = 311

cf = cumulative frequency of the class preceding median class = 289

f = frequency of the median class = 102

h = class interval of the median class = 10

Substituting these values in the formula of median, we get

⇒

⇒ Median = 60 + 2.157

⇒ Median = 62.157

Thus, the median of data is 62.157.

Given is, median of the data = 32

Sum of frequency, N = 100

To find: x and y

Let us show cumulative frequency in a table.

Here, we have got

N = 75 + x + y

But, N = 100

⇒ 75 + x + y = 100

⇒ x + y = 100 – 75

⇒ x + y = 25 …(i)

For median:

We have, total frequency, N = 100

N/2 = 100/2 = 50

∵ median = 32

It lies in the interval 30 – 40.

Thus, median class = 30 – 40

Median is given by

Where,

L = Lower class limit of median class = 30

N/2 = 50

cf = cumulative frequency of the class preceding median class = 35 + x

f = frequency of the median class = 30

h = class interval of the median class = 10

Substituting these values in the formula of median, we get

⇒

⇒

⇒

⇒

⇒ 96 = 105 – x

⇒ x = 105 – 96

⇒ x = 9

Substituting x = 9 in equation (i),

x + y = 25

⇒ 9 + y = 25

⇒ y = 25 – 9

⇒ y = 16

Thus, x = 9 and y = 16.

We have been given data of profit of each shop in class interval and data of number of shops.

Let us create greater than type cumulative frequency distribution table.

Taking Profit of each shop as x-axis and cumulative frequency as y-axis.

Thus plotting the points on a graph, we get

We have been given data of weight in less than type and data of number of students.

We have

Taking weight as x-axis and number of students as y-axis.

Thus plotting the points on a graph, we get

Now we need to plot median in the graph and then verify it using the formula.

So, note that total number of students are 35.

⇒ N = 35

Find N/2.

⇒

Draw a line parallel to x axis, passing through 17.2 on y-axis intersecting the less than ogive.

We have graphically,

We can notice from the graph, the line intersecting the ogive touches the y-axis at 46.7.

Now, let us solve theoretically,

Let us make a table showing frequencies and class intervals.

Since,

Observe, cf = 28 is just greater than 17.5.

Thus, median class = 46 – 48

Median is given by

Where,

L = Lower class limit of median class = 46

N/2 = 17.5

cf = cumulative frequency of the class preceding median class = 12

f = frequency of the median class = 16

h = class interval of the median class = 2

Substituting these values in the formula of median, we get

⇒

⇒ Median = 46 + 0.6875

⇒ Median = 46.6875

Thus, the median of data is 46.6875 and it is verified.

We have been given class intervals and frequency.

Let us create greater than type of cumulative frequency distribution table.

Taking Class as x-axis and cumulative frequency as y-axis.

Thus plotting the points on a graph, we get

Thus, this is the more than type Ogive on graph paper.

We have been given class intervals and frequency.

Let us create greater than type as well as less than type of cumulative frequency distribution table.

Taking Class as x-axis and cumulative frequency as y-axis separately for greater than type as well as less than type.

Thus plotting the points on a graph, we get

The median is given by intersection point of less-than type and more-than type ogives and is represented on x-axis.

Note here, the arrow points at approximately 139.

Even if we verify by using the formula, we’d get an answer around 139.

Since,

Observe in the less than type cumulative frequency, cf = 26 is just greater than 25.

Thus, median class = 120 – 140

Median is given by

Where,

L = Lower class limit of median class = 120

N/2 = 25

cf = cumulative frequency of the class preceding median class = 12

f = frequency of the median class = 14

h = class interval of the median class = 20

Substituting these values in the formula of median, we get

⇒

⇒ Median = 120 + 18.57

⇒ Median = 138.57 ∼ 139

Thus, the median of data is 138.57 and it is verified too.

Mode is just the set of data value that appears most often.

Let us arrange the data values in ascending order so that observation can be made easily.

10, 11, 15, 15, 15, 15, 16, 16, 16, 17, 17, 17, 17, 18, 18, 19

Note the appearances now,

10 appears → 1 time

11 appears → 1 time

15 appears → 4 times

16 appears → 3 times

17 appears → 4 times

18 appears → 2 times

19 appears → 1 time

15 and 17 both appears most times, that is, 4 times.

Thus, mode is 15 and 17.

Mode is just the set of data values that appears most often.

Let us arrange the data values in ascending order so that observation can be made easily.

130, 130, 130, 130, 130, 131, 131, 131, 131, 131, 131, 131, 132, 132, 132, 132, 133, 133, 133, 133, 134, 134, 135, 135, 135, 135

Note the appearances now,

130 appears → 5 times

131 appears → 7 times

132 appears → 4 times

133 appears → 4 times

134 appears → 2 times

135 appears → 4 times

⇒ 131 appears most of the times, that is, 7 times.

Thus, mode is 131.

Since, mode is the set of data values that appears most often in the data.

(i). Arranging the data values in ascending order to make observation easier, we get

2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 8

Note the appearances now,

1 appears → 1 time

3 appears → 4 times

4 appears → 12 times

5 appears → 9 times

6 appears → 2 times

7 appears → 1 time

8 appears → 1 time

⇒ 4 appears most of the times, that is, 12 times.

Thus, mode is 4.

(ii). Arranging the data values in ascending order to make observation easier, we get

2, 3, 8, 8, 10, 10, 10, 10, 10, 10, 11, 11, 14, 15, 15, 15, 15, 15, 17, 17, 18, 18, 19, 19

Note the appearances now,

2 appears → 1 time

3 appears → 1 time

8 appears → 2 times

10 appears → 6 times

11 appears → 2 times

14 appears → 1 time

15 appears → 5 times

17 appears → 2 times

18 appears → 2 times

19 appears → 2 times

⇒ 10 appears most of the time, that is, 6 times.

Thus, mode is 10.

Since, mode is just the data value/s appearing most often in given set of data values.

We just need to search for the size which has the greatest frequency. Frequency is just the number of occurrence of the data value.

Here,

2 appears → 3 times

3 appears → 4 times

4 appears → 5 times

5 appears → 3 times

6 appears → 5 times

7 appears → 4 times

8 appears → 3 times

9 appears → 2 times

⇒ 4 and 5 appears most of the times, that is, 5 times.

Thus, mode is 4 and 5.

This is a grouped frequency distribution table.

Notice the greatest number of examinees is 75.

⇒ Modal class = 18 – 20

Mode is given by

Where,

L = Lower class limit of the modal class = 18

h = class interval of the modal class = 2

f₁ = frequency of the modal class = 75

f₀ = frequency of the class preceding the modal class = 45

f₂ = frequency of the class succeeding the modal class = 38

Substituting values in the formula of mode,

⇒

⇒

⇒ Mode = 18 + 0.896

⇒ Mode = 18.896

Thus, mode is 18.896.

This is a grouped frequency distribution table.

Notice the greatest number of students is 22.

⇒ Modal class = 20 – 25

Mode is given by

Where,

L = Lower class limit of the modal class = 20

h = class interval of the modal class = 5

f₁ = frequency of the modal class = 22

f₀ = frequency of the class preceding the modal class = 16

f₂ = frequency of the class succeeding the modal class = 11

Substituting values in the formula of mode,

⇒

⇒

⇒ Mode = 20 + 1.76

⇒ Mode = 21.76

Thus, mode is 21.76.

This is a grouped frequency distribution table.

Notice the greatest frequency is 28.

⇒ Modal class = 15 – 20

Mode is given by

Where,

L = Lower class limit of the modal class = 15

h = class interval of the modal class = 5

f₁ = frequency of the modal class = 28

f₀ = frequency of the class preceding the modal class = 18

f₂ = frequency of the class succeeding the modal class = 17

Substituting values in the formula of mode,

⇒

⇒

⇒ Mode = 15 + 2.38

⇒ Mode = 17.38

Thus, mode is 17.38.

This is a grouped frequency distribution table which is inclusive type.

We need to convert this inclusive type of data into exclusive type of data to find mode of these data values.

To convert to exclusive type of data, subtract 0.5 from the lower class of the interval and add 0.5 to the upper class of the interval.

Notice the greatest frequency is 32.

⇒ Modal class = 74.5 – 84.5

Mode is given by

Where,

L = Lower class limit of the modal class = 74.5

h = class interval of the modal class = 10

f₁ = frequency of the modal class = 32

f₀ = frequency of the class preceding the modal class = 19

f₂ = frequency of the class succeeding the modal class = 12

Substituting values in the formula of mode,

⇒

⇒

⇒ Mode = 74.5 + 3.94

⇒ Mode = 78.44

Thus, mode is 78.44.

Frequency curve is a smooth curve which corresponds to the limiting case of a histogram computed for a frequency distribution of a continuous distribution as the number of data points becomes very large.

Frequency polygons are a graphical device for understanding the shapes of distributions. They serve the same purpose as histograms, but are especially helpful for comparing sets of data.

A histogram is an accurate representation of the distribution of numerical data. It is an estimate of the probability distribution of a continuous variable (quantitative variable).

An ogive, sometimes called a cumulative frequency polygon, is a type of frequency polygon that shows cumulative frequencies. Ogives do look similar to frequency polygons, which we saw earlier. The most important difference between them is that an ogive is a plot of cumulative values, whereas a frequency polygon is a plot of the values themselves.

⇒ Ogives can easily estimate median of a given frequency distribution since they are plots of given class and frequency and median can be estimated on x-axis.

Mean is given by adding the data values or variables and dividing them by the number of data values or variables.

So,

⇒

⇒ 54 = 35 + x + y

⇒ 54 – 35 = x + y

⇒ x + y = 19

Thus, option (B) is correct.

First, let us find out median of the original set of data.

To find its median, arrange the data into ascending order. We get

30, 34, 35, 36, 37, 38, 39, 40

(Since, it was already arranged in ascending order)

Total number of values, n = 8

Since n is even,

⇒

⇒ Median = The mean of 4^th term and 5^th term

⇒ Median = The mean of 36 and 37

⇒

⇒

⇒ Median = 36.5 …(i)

And if according to the question, 35 is removed then we have the data,

30, 34, 36, 37, 38, 39, 40

Total number of values, n = 7

Since n is odd,

⇒

⇒

⇒ Median = 4^th term

⇒ Median = 37 …(ii)

Subtracting equations (ii) and (i), we get

Difference = 37 – 36.5

⇒ Difference = 0.5

Thus, option (D) is correct.

Mode of the data is the value that occurs most of the time in data

Now we can see from data given to us that 16 occurs 2 times and 15 occurs 2 times, rest of the values occur only 1 time

So, for 15 to be the mode of data, 15 should occur three times and this can happen only when x = 15

So, x = 15

We have arranged data,

8, 9, 12, 17, x + 2, x + 4, 30, 31, 34, 39

Median = 15

Total number of data values, n = 10

Since, n is even, median will be given by

⇒

⇒ Median = The mean of 5^th term and 6^th term

⇒ Median = The mean of (x + 2) and (x + 4)

⇒

⇒

⇒ 30 = 2x + 6

⇒ 2x = 30 – 6

⇒ 2x = 24

⇒

⇒ x = 12

(i). The statement is false.

To check this, we need to find mode of 2, 3, 9, 10, 9, 3, 9.

Count the data values:

2 appears → 1 time

3 appears → 2 times

9 appears → 3 times

10 appears → 1 time

⇒ 9 appears most of the times, that is, 3 times.

Thus, mode is 9.

Hence, the statement given is false.

(ii). The statement is false.

To check this, we need to find median of 3, 14, 18, 20, 5.

Arrange the data into ascending order. We get

3, 5, 14, 18, 20

Total number of values, n = 5

Since n is odd, median is given by

⇒

⇒

⇒ Median = 3^rd term

Thus, median is 14.

Hence, the statement is false.

(i). Mean, median and mode are measures of central tendency.

A central tendency (or measure of central tendency) is a central or typical value for a probability distribution.

Mean gives us the average of given set of values or variables, which is the central value in a set of values.

Median is a value separating higher half from the lower half of the data, which gives us the central measure.

Mode is the value in the data occurring most often, gives an indication of the central value in the data.

(ii). If mean of is mean of is .

Given that, mean of x₁, x₂, x₃, …, x_n is .

⇒

⇒

⇒ x₁ + x₂ + x₃ + …+ x_n = n …(1)

Then, mean of ax₁, ax₂, ax₃, …, ax_n is given by

⇒

⇒

⇒ Mean = a

(iii). At the time of finding arithmetic mean, the lengths of all classes are not necessarily equal.

It is not necessary for class intervals to be equal but it would be better if class interval size is smaller. But at the time of finding arithmetic mean, equal or unequal class interval can yield result.

Here, we need to find the median class and modal class.

Let us find median class.

Let us create a table.

Here, total sum of frequency, N = 77

⇒

⇒

Note in the table that, cf = 42 is just greater than 38.5.

⇒ Median class = 125 – 145

Now, let us find modal class.

For modal class, just check the greatest frequency.

Here, the greatest frequency = 20

⇒ Modal class = 125 – 145

Upper class limit of median class = 145

Lower class limit of modal class = 125

Difference = Upper class limit of median class – Lower class limit of modal class

⇒ Difference = 145 – 125

⇒ Difference = 20

Thus, answer is 20.

To find modal class, find the greatest frequency.

⇒ Greatest frequency = 71

So, corresponding to this value of frequency, we get class = 14.2 – 14.6

⇒ Modal class = 14.2 – 14.6

So, upper class limit of modal class = 14.6

Lower class limit of modal class = 14.2

Difference = Upper class limit of modal class – Lower class limit of modal class

⇒ Difference = 14.6 – 14.2

⇒ Difference = 0.4

Thus, difference is 0.4.

Given: Mean = 8.1

∑f_ix_i = 132 + 5k

∑f_i = 20

To find k, we need to use the direct method formula of mean.

⇒

⇒

⇒ 162 = 132 + 5k

⇒ 5k = 162 – 132

⇒ 5k = 30

⇒

⇒ k = 6

Thus, k = 6.

Given that, …(i)

∑f_iu_i = 20

∑f_i = 100

From the step deviation method, we have the relation of deviation (u_i) and assumed mean (A).

Comparing it with equation (i),

A = 25

h = 10

Using the formula of step deviation method, we have

⇒

⇒ Mean = 25 + 2

⇒ Mean = 27

To find modal class, we need to find frequencies of the data given.

Note the greatest frequency in the data.

So, greatest frequency = 30

Thus, the corresponding class is 30 – 40.

⇒ Modal class = 30 – 40

Thus, mean is 27 and modal class is 30 – 40.