WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode

WBBSE Class 10 Median and Mode Overview

The median or the mid-values is another measure of central tendency.

If the values of the variable x be arranged in ascending or descending order, then the value in the middle position of the arrangement is called the median.

There are equal number of values above and after the median.

Let x₁,x₂,……,x_n be n-values of the variable x. If we arrange the values in ascending order either, we get, \(x_{(1)}, x_{(2)}, \ldots \ldots, x_{(n)}, \text { i.e., } x_{(1)} \leq x_{(2)} \ldots \ldots \leq x_{(n)}\)

Amongst the values, there may occur some values which are equal to each other. But not all are equal.

Now, if n is an odd number, then the \(\left(\frac{n+1}{2}\right)\)-th value is the median or mid-value.

WBBSE Solutions for Class 10 Maths

∴ the median = \(\mathrm{Me}=x_{\left(\frac{n+1}{2}\right)}\)…….(1)

If n is an even number, then we shall get no unique median of the values of x.

Then the two values \(\left(\frac{n}{2}\right)-\text { th and }\left(\frac{n}{2}+1\right)\)-th values are denoted as the medians of the values i.e,

here, \(x_{\left(\frac{n}{2}\right)} \leq \mathrm{Me} \leq x_{\left(\frac{n}{2}+1\right)}\)……(2)

For unique median we generally find the average of the two values \(\left(\frac{n}{2}\right)-\text { th and }\left(\frac{n}{2}+1\right)\)-th values, i.e.,

\(\mathrm{Me}=\frac{x_{\left(\frac{n}{2}\right)}+x_{\left(\frac{n}{2}+1\right)}}{2}\)

Statistics Chapter 3 Median And Mode Examples

Examples of Median Calculation

Example 1. Find the median of the values in 15, 19, and 20.

Solution:

Given:

15, 19, and 20

Here, arranging the values in ascending order, we get, 15, 19, 20.

Here, the number of numbers is 3.

∴ the \(\left(\frac{3+1}{2}\right)\)-th i.e. the second value is the required median.

∴ Median = 19 quintals.

Example 2. Find the median of the values 17, 25, 30, 32, 28, 24, 20, 18, 16, 10 Arranging the values in ascending order we get 10, 16, 17, 18, 20, 24, 25, 28, 30, 32.

Solution:

Given:

10, 16, 17, 18, 20, 24, 25, 28, 30, 32

Here, the number of numbers is 10, which is an even number.

So, there are two mid-values, the \(\left(\frac{n}{2}\right) \text {-th and }\left(\frac{n}{2}+1\right)\)-th values, i,e., the 5th and 6th values are mid-values.

Thus, the two mid-values are 20 and 24.

∴ the required median = \(\frac{20+24}{2}\) = 22

∴ the required median = 22 kilometers/hour.

Example 3. Find the median of the values are given in below table

Solution: Here, arranging the values in ascending order, we get,

Here, the number of values is 41 and so the median is the \(\left(\frac{41+1}{2}\right)\)th i.e- 21st value.

Here, the 21st value is 24.

∴ the required median = 24.

Example 4. Find the median of clothes 2 shares of prices Rs.30 each, 3 sharees of process Rs.35 each, and 5 sharees of price Rs.45 each is given.

Solution:

Given:

clothes 2 shares of prices Rs.30 each, 3 sharees of process Rs.35 each, and 5 sharees of price Rs.45 each is given.

Here, amongst 10 clothes, the price of 2 is Rs. 30, the price of 3 is Rs. 35 and the price of 5 is Rs. 45.

Now, arranging the prices in ascending order we get 30,30,35,3.5,35,45,45,45,45,45.

Here, the total number of clothes is 10 which is an even number, the mid-values are 2 in number, i.e. the \(\left(\frac{10}{2}\right)\)th and \(\left(\frac{10}{2}+1\right)\)th values are the mid-values, i.e., the 5th and 6th values are the mid-values.

Here the 5th value = 35 and the 6th value = 45.

∴ the median = \(\frac{35+45}{2}\) = 40.

∴ the required median = Rs. 40.

Example 5. Find the median of the telephone calls given in the below table

Solution: Here, the values of the frequency distributions are already arranged in ascending order.

In this case, 0 occurs 2 times, and 1 occurs 5 times. 2 occurs 6 times, etc.

Now, preparing the cumulative frequency distribution table, we get,

Table: Finding of the median of the telephone calls

From the table, we see that the number of values up to 3 is 24, and up to 4 it is 36.

Here, the total number of values is 60.

So, the average of the 30th and 31st values is to be taken as the median.

Since every value from the 25th to 36th values is 4, so both the 30th and 31st values is 4. ∴ the median = 4.

Example 6. Find the medians of the diameters Here, also we have to prepare the following cumulative frequency distribution.

Solution:

Determination of the median of the diameters

Here, 25 values are each 45, 28 values are each 46, etc.

From the cumulative frequency, we see that the number of values up to 46 is 53, up to 47 is 85.

Here, the total number of values is 150. So, the 75th and 76th values are to be taken as the median.

Since from 54th to 85th values are all.47, so both 75th and 76th values are 47.

∴ the required median = 47 mm

Example 7. Find the median of heights in the given table.

Solution: For continuous variables, at first we have to find the class boundaries and their corresponding cumulative frequencies.

The value for which the cumulative frequency is \(\frac{n}{2}\) is assumed to be the median.

Since here the frequencies of each class interval are given and the frequencies of different values are not available, it is impossible to determine the correct value of the median.

Frequencies of the different class intervals are equally distributed assuming this principle, we can find the required median. To do so, let us prepare the following tables:

Here, \(\frac{n}{2}\) = \(\frac{100}{2}\) = 50.

From the cumulative frequencies we see that up to 159.5, the total frequency is 33, and up to 164.5, it is 63.

The value 50 lies in between 33 and 63. So, the median will lie in between the class boundary 159.5 to 164.5.

The frequency of the two class boundaries is 159.5 and 164.5.

The frequency of the two class boundaries 159.5 and 164.5 is 30 and the class length is 5.

So, by the method of proportional part, we get,

\(\frac{\text { Median }- \text { Lower class boundary }}{\text { Upper class boundary – Lower class boundary }}\)=\(\frac{50-33}{63-33}\)

 

or, \(\frac{\text { Median }-159 \cdot 5}{164 \cdot 5-159 \cdot 5}=\frac{50-33}{63-33} \text { or, Median }-159 \cdot 5=\frac{17}{30} \times 5\)

or, Median – 159.5 = 2.83 or, Median = (159.5 + 2.83) cm = 162.33 cm

In general, if x_i be the lower boundaries of the class-interval jn which the median belongs to f₀ be the frequency of that class interval F be the cumulative frequency upto x_i, n be the total frequency, and c be the class length, then

Median, \(\mathrm{Me}=x_l+\frac{\frac{n}{2}-\mathrm{F}}{f_0} \times c\)……..(4)

Otherwise, for the value of \(\frac{n}{2}\), the value of the x-coordinate of the ogive (less or greater than type) is taken as the median.

Example 8. Find the median of the numbers frequency distribution of the marks obtained by 76 students.

Solution: We have to prepare the following table:

Determination of the median of the numbers:

Here, total frequency = 76 and \(\frac{n}{2}\) = 38.

38 is in between 18 and 50. So, the median lie iu between 20.50 and 30.5.

Here x₁= 20.5, F = 18, f0 = 32, c =10

∴ Median = \(x_l+\frac{\frac{n}{2}-\mathrm{F}}{f_0} \times c=20 \cdot 5+\frac{38-18}{32} \times 10\)

= \(20 \cdot 5+\frac{20}{32} \times 10=20 \cdot 5+6 \cdot 25=26 \cdot 75\)

∴ the required median = 26.75

Example 9. The frequency distribution of the incomes of 75 employees of the industry are given below. Find the median.

Solution: Determination of the median of the incomes :

Here, the total frequency n = 75 and \(\frac{n}{2}\) = 37.5 lies in between 20 and 45.

So, the median lies in the class interval 250- 500.

Here, x₁ = 250, F = 20, f₀ = 25, c = 250.

∴ Median = \(250+\frac{37 \cdot 5-20}{25} \times 250=250+\frac{17 \cdot 5}{25} \times 250=250+175=425\)

∴ the required median = Rs 425.

[Note: In the above example the first and last class intervals are open class-interval.]

Mode:

Mode is another measure of the central tendency. The value of the variable x, which occurs the highest times than all other values of the variable x in the given data is known as the mode of that data.

We briefly denote it by MQ.

For discrete variables, the greatest frequency of a frequency distribution is usually called its mode.

For a continuous variable, the value which have the highest frequency density is known as the mode.

Sometimes mere may occur values more than once, which have, the highest frequency or frequency density.

In these cases, it is impossible to determine the mode uniquely. Here, the variable or the given frequency distribution is said to be of several modes.

Statistics Chapter 3 Median And Mode Mode Examples

Finding Mode in Frequency Distribution

Example 1. Find the mode of the data is 15 quintals, 19 quintals, and 20 quintals are given.

Solution:

Given:

Data is 15 quintals, 19 quintals, and 20 quintals .

Here, the three given Values of the variable are 15 quintals, 19 quintals, and 20 quintals, which occur once each in the data.

So, here it is impossible, to determine mode uniquely.

Example 2. Find the mode of velocities 17, 25, 30, 32, 28, 24, 20, 18, 16, and 10 are given.

Solution:

Given:

velocities 17, 25, 30, 32, 28, 24, 20, 18, 16, and 10.

Here also, each value occurs once in the given data.

So, here also it is impossible to determine the mode uniquely.

Example 3. Find the mode of the numbers are given below:

Solution: Observe that each of the numbers 17, 21, 23, and 25 occur in 3 times each.

So, here also it is impossible to determine the mode uniquely.

Example 4. Find the mode of the selling prices 115, 98, 102, 126, 85, 91, and 107 are given.

Solution:

Given:

The selling prices 115, 98, 102, 126, 85, 91, and 107

Here, the 7 different values each occurs once in the given data.

So, here also it is impossible to find the mode uniquely.

Example 5. Find the mode of the prices of clothes 2 shares of price Rs.30 each, 3 shares of price Rs.35 each, and 5 shares of price Rs. 45 each.

Solution: Here, Rs.30 occurs 2 times, Rs. 35 occurs 3 times, and Rs. 45 occurs 5 times, i.e., Rs. 45 occurs the highest times.

So, the required mode = Rs 45.

Mode Calculation with Grouped Data

Example 6. Find the mode of the telephone calls given per minute for any hour are given below:

Solution: Here, telephone calls are discrete variable.

We can easily find from the frequency distribution that the frequency of 4 is 12 and it is the highest time to be occurred.

∴ The mode = 4.

Example 7. Find the mode of the frequency distribution of diameters.

Solution: Here, amongst the 5 different values of the variable, the frequency of 4 mm is 35 and it is the highest.

∴The required mode = 48 mm

Example 8. Find the mode of the frequency distribution of the height (in cm) of 100 students.

Solution: Here, height is a continuous variable.

It is very difficult to determine the mode of a continuous variable.

The easiest way, to determine the mode of a continuous variable, is to construct a histogram of the variable and then a similar or congruent frequency line along with the histogram.

The value of the variable for which the frequency line is the highest is its mode.

In the following, here is an imaginary histogram and its frequency line and so that the mode is determined.

 

To determine this frequency line congruently is not an easy matter.

Roughly we may take the mid-value of the class interval which have the highest frequency density as the mode.

i.e., \(\mathrm{M}_0 \cong \frac{x_u+x_l}{2}=x_0 \text { (let) }\)…..(5)

Again, let the class intervals are equal and each of them be c.

If the highest frequency be f₀, the frequency of class interval of the previous class interval be f_(-1) and the frequency of the class interval of the next class interval be f₁.

Now, it is possible to take x₀ as the mode if f₀– f_(-1) and f₀– f₁ are equal.

If f₀– f_(-1) is greater than f₀– f₁, the mode is nearer to x_i and if it is alternative, the mode is nearer to x₁.

According to the proportional law,

\(\frac{\mathrm{M}_0-x_l}{x_u-\mathrm{M}_0}=\frac{f_0-f_{-1}}{f_0-f_1}\)

 

or, \(\frac{\mathrm{M}_0-x_l}{x_u-x_l}=\frac{f_0-f_{-1}}{2 f_0-f_1-f_{-1}}\)

or, \(\mathrm{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_1-f_{-1}} \times c \quad\left[because x_u-x_l=c\right]\)

\( \bar{x}-M_0=3\left(\bar{x}-M_e\right)\)

 

or, \(\mathrm{M}_0=3 \mathrm{M}_e-2 \bar{x}\)…….(6)

There is no proof of the above relation, but in maximum cases, it is seen that the relation (6) is roughly satisfied.

Using the formula (4), we get

\(\text { Mode }=\mathrm{M}_0=159 \cdot 5+\frac{30-22}{2 \times 30-22-24} \times 5, \text { where } \mathrm{f}_0=30, \mathrm{f}_{-1}=22, \mathrm{f}_1=24, \mathrm{c}=5\)

 

and \(\mathrm{x}_1=159 \cdot 5 \text { and } \mathrm{x}_{\mathrm{u}}\)

= 164.5

= \(\left(159 \cdot 5+\frac{8}{14} \times 5\right) \mathrm{cm}\) =(159.5 + 2.86)= 162.36 cm

Again, from the formula (6), we get,

Mode = \(\mathrm{M}_0=3 \times \mathrm{Me}-2 \bar{x}=3 \times 162 \cdot 33-2 \times 162 \cdot 40,\)

[where \(\bar{x}\) =162.40 and \(\mathrm{M}_e=162 \cdot 33\)]

Also, from the formula (6), we get,

Mode = 3 x M_e – 2\(\bar{x}\)

= 3 x 162.33 – 2 x 162 = 486.99 – 324.80 = 162.19

∴ the required mode = 162.19 cm.

Example 9. Find the mode of the marks frequency distribution obtained by 76 students.

Solution: Here, the mode lies in the interval 20.5- 30.5, since its frequency density is the highest.

∴ x₁= 20.5, x_u = 30.5, c = 10, f₀ = 32, f_-1 = 14 and f₁ = 18

∴ By formula, the mode = 20.5 + \(\frac{32-14}{2 \times 32-14-18}\) x 10 = 20.5 + \(\frac{18}{32}\) x 10

= 20.5 + 5.62 = 26.12 (approx.)

By another formula, M_e = 26.75 and \(\bar{x}\) = 27.08,

Mode = 3 x M_e– 2\(\bar{x}\) = 3 x 26.75 – 2 x 27.08 = 80.25 – 54.16 = 26.09

Applications of Median and Mode in Real Life

Comparison of Mean, Median, and Mode

Mean, median add mode all three are measures of central tendency.

All three measures are easy, comprehensible, and have comprehensible significance.

But to the general public, Mean is very well known and is widely used.

All three measures can easily be determined, although for the continuous variable, it is difficult to determine to die mode and we generally become satisfied by its approximate value Among the three measurements, only the mean is dependable directly on all the obtained values of the variable because in the case of other two measurements, we can change the other values of the measurements by taking the measurement unchangeable.

If the mid-values and mode remain unchanged, then the median and mode also remain unchanged.

However, you have seen that the Mean has some easy properties, so that is it easy to use. Median and mode have no such properties.

There are some special areas where the Use of Mean is impossible and improper. Let amongst 7 person income of 6 persons are in between Rs. 500 and Rs. 600 and the income of the 7th person is Rs. 2000.

If we determine the Mean of it we shall see that it is more than Rs.-600 and it can never be the measure of central tendency.

So, if some of the values be more greater or smaller compared to the other values, then it is improper to determine the mean of the variable.

Here the first and last class intervals are open. As a result, determining the mid-values of that two class intervals are impossible. So, it is also impossible to find out the mean.

In this case, it is easier to find out the median. So if the first and last or both the class intervals are open, then it is reasonable to determine the median and mode as the central tendency.

Example 10. The median and mode of the 61 values of a variable are 17 and 19 respectively. Later on, it is seen that one of the values is wrongly taken as 13, the correct value of which is 16. Find the correct values of median and mode.

Solution:

Given:

The median and mode of the 61 values of a variable are 17 and 19 respectively. Later on, it is seen that one of the values is wrongly taken as 13, the correct value of which is 16.

If we arrange 61 values of the variable in ascending or descending order, then by definition, the 1st i.e., the 31st value is the median. The wrong number is 13.

So, if we arrange the values in ascending order, then its order of value will be below 31. Again, instead of 13, if 16 is taken, then the value of the mean will be also below the 31st value.

So, the median = 31st value will remain the same. Mode is the value of the variable with the highest frequency. Therefore, if any other values of the variable be changed, then the mode will remain unchanged.

So, the mode remains 19.

Word Problems on Median and Mode

Example 11. The mean of the frequency distribution of a variable is 25 and its median is 23. Find the value of its mode.

Solution:

Given:

The mean of the frequency distribution of a variable is 25 and its median is 23.

We know that the relationship between mean, median, and mode is given by Mean- Mode = 3 (Mean – Median)

∴ Mode = 3 x Median- 2 x Mean = 3 x 23- 2 x 25 = 69- 50 = 19

∴ the required mode =19.

Example 12. Find the Median and Mode of the variable from the following frequency distribution table:

Solution: Here, the class ranges are not equal. There will be no problem in determining the median, but to find the mode, the class ranges must be made equal.

Let us draw the following table of cumulative frequency distribution for the median.

We know that median, \(\mathrm{M} e=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f} \times c\)

Here, \(\frac{N}{2}\) = \(\frac{120}{2}\) = 60

∴ the median lies from 20.5 to 25.5, ∴ x_i= 20.5, F = 59 and f = 19

c = the range of the class interval in which median lie = 25.5- 20.5 = 5

∴ the required median, Me = 20.5 + \(\frac{60-59}{1.9}\) = 20.5 + \(\frac{5}{19}\) = 20.5 + 0.263 = 20.76

Now, rearranging the above frequency distribution table by equalising the class ranges we get,

We know that Mode, \(\mathbf{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, \(x_l=10 \cdot 5 ; f_0=52 ; f_{-1}=7 ; f_1=34 \text { and } c=10\)

∴ the required mode = 10.5 + \(\frac{52-7}{2 \times 52-7-34}\) x 10

= 10.5 + \(\frac{45}{104-41}\) x 10 = 10.5 + \(\frac{450}{63}\) = 10.5 + 7.143 = 17.643

Example 13. The monthly incomes (in rupees) of 70 persons are given in the following frequency distribution table:

Find the mode of the monthly income.

Solution: Let us draw a frequency distribution table:

According to the formula, \(\mathbf{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, \(x_l=299 \cdot 5, f_0=21, f_{-1}=15, f_1=7 \text { and } \mathrm{c}=100\)

∴ the required mode, \(M_0=299 \cdot 5+\frac{21-15}{2 \times 21-15-7} \times 100\)

= \(299 \cdot 5+\frac{6}{42-22} \times 100=299 \cdot 5+\frac{6}{20} \times 100=299 \cdot 5+30=329 \cdot 50\)

∴ the required Mode = Rs. 329.50

Example 14. The frequency distribution table of the prices of some houses (in Rs. 1000) is given in the following. In the table, the number of houses costing from Rs. 30,000 to Rs. 40,000. If the median of the prices of houses be Rs. 40073, then find the total number of houses.

Solution: Let us draw a less-than-type cumulative frequency distribution table as follows

The units when changed in Rs. 1000, then the median is 40,073 and it belongs to the class intervals 40-50.

Now, the median, \(\mathrm{Me}=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_m} \times c\)

Here, \(\mathrm{Me}=40 \cdot 073, x_l=40 ; \frac{\mathrm{N}}{2}=\frac{497+f}{2} ; \mathrm{F}=196+f ; c=10 \text { and } f_m=137\)

Putting these values in the formula,

\(40 \cdot 073=40+\frac{\frac{497+f}{2}-(196+f)}{137} \times 10\)

 

or, \(\frac{40 \cdot 073-40}{10} \times 137=\frac{497}{2}+\frac{f}{2}-196-f\)

or, 1.0001 = 248.5- 196 – \(\frac{f}{2}\)

or, 1.0001 = 52.5-\(\frac{f}{2}\) or, \(\frac{f}{2}\) = 52.5 – 1.001 or, \(\frac{f}{2}\) = 51.499 or, f = 102.9998 = 103

∴ Total number of houses = N = 497 + 103 = 600

Example 15. The heights (in cm.) of 180 aged male Indians are given in the following frequency distribution table. Two of its frequencies of two class intervals are not given. The median of the heights is 163.133 Find the unknown two frequencies.

Solution: Let us draw a less-than-type cumulative frequency table from the given table

Here, Median Me = 163.133. It lies in the class interval 161.5- 16.7.5

We know, total frequency, N = 180 = 98 + f₁ + f₂

∴ f₁ + f₂ = 180- 98 or, f₁ + f₂ = 82 ……(1)

Again, by the formula, \(\mathrm{M} e=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_m} \times c\)

Here, Me = 163.133, x₁ = 161.5, \(\frac{N}{2}\) = \(\frac{180}{2}\) = 90, F = 28 + f₁, and c = 6 (cm.), f_m = 60

Substituting these values in the formula we get,

163.133 = 161.5 + \(\frac{90-28-f_1}{60}\) x 6

or, 163-133 = 161.5 + \(\frac{62-f_1}{10}\)

or, 163.133-161.5 = \(\frac{62-f_1}{10}\)

or, 1.633 x 10 = 62- f₁ or, f₁ = 62- 16.33 = 45.67 = 46

Now, putting the value of f₁ in (1) we get, 46 + f₂ = 82 or, f₂ = 82- 46 = 36

∴ f₁ = 46 and f₂ = 36

Visual Representation of Median and Mode

Example 16. The frequency distribution table of the lengths (in cm.) of the vertex of which is given below. The frequency of the class-interval 9.7- 10.4 in this table is unknown. If the mode of the lengths be 9.917 cm, then find the unknown frequency.

Solution: It is seen that the mode lies in the interval 9.7- 10.4

By the formula, Mode, \(\mathbf{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, M₀ = 9.917; x,= 9.65; f_x = 33; / = 30 and c = 0.8.

9.917 = 9.65 + \(\frac{f-33}{2 f-33-30}\) x 0.8

or, \(\frac{9.917-9.65}{0.8}=\frac{f-33}{2 f-63} or, \frac{0 \cdot 267}{0.8}=\frac{f-33}{2 f-63}\)

or, \(0.334=\frac{f-33}{2 f-63} or, 0.668 f-21 \cdot 042=f-33\)

or, \((1-0.668) \cdot f=33-21.042 or, 0.332 f=11.958\)

or, \(f=\frac{11 \cdot 958}{0 \cdot 332}=36 \cdot 01 \approx 36\).

Hence, the unknown frequency = 36.

Example 17. The frequency distribution of the heights (in inches) of 200 persons is given below. If the mode of the heights be 61.167 inches, find the unknown frequencies x and y.

Solution: Here, the mid-points of the class intervals are given.

The class boundaries are 57.5- 58.5, 58.5- 59.5, 59.5-60.5,……etc.

Mode, M₀ = 61.167 inches, it is in the class interval 60.5- 61.5.

Now, by the formula of Mode, \(\dot{\mathrm{M}}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, \(\mathrm{M}_0=61: 167, x_l=60 \cdot 5, f_0=44, f_{-1}=x, f_1=y \text { and } c=1\)

Substituting these values in the formula, we get,

\(61 \cdot 167=60 \cdot 5+\frac{44-x}{88-x-y} \times 1\)

 

or, \(61 \cdot 167-60 \cdot 5=\frac{44-x}{88-x-y}\)

or, 0.667 = \(\frac{44-x}{88-x-y}\)

or, 58.696 – 0.667x – 0.667y = 44 – x or, 58.696 – 44 = 0.667y -0.333x

or, 14.696 = 0.667 y- 0.333x….. (1)

Sum of the frequencies = 130 + x+y

∴ 200= 130 + x + y or,70 = x + y……(2)

Solving (1) and (2) we get, x = 32 and y = 38

∴ x = 32 and y = 38

The unknown frequencies x and y 32 and 38

Example 18. Find the mode and median of the following frequency distribution table:

Solution: Let Us draw a cumulative frequency distribution table of the above-given table

By the formula of median \(\mathrm{M} e=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_0} \times c\)

Here, \(\frac{N}{2}\) = \(\frac{100}{2}\) = 50

∴ the median lies in between 60 and 65.

Here, x₁ = 60, F = 44 and f₀ = 30 and c = 65- 60 = 5

∴ the required median, \(\mathrm{M} e=60+\frac{50-44}{30} \times 5=60+\frac{6 \times 5}{30}\) = 60 + 1 = 61

Also, by the formula of mode, we have,

\(\mathrm{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

 

Here, \(x_l=60, f_0=30, f_{-1}=20, f_1=15\) and c = 5

Putting these values in the above formula, we get,

∴ the required mode, \(\mathrm{M}_0=60+\frac{30-20}{2 \times 30-20-15} \times 5\)

= \(60+\frac{10 \times 5}{60-35}=60+\frac{50}{25}\) = 60+ 2 = 62.

∴ the required mode = 62 cm and the required median = 61 cm.

Example 19. The frequency distribution of the marks obtained by 50 students in Math is given below:

Find the median of the marks approximated to two decimal points.

Solution: Let us draw a cumulative frequency distribution table (less than type) of the above-given data.

By the formula of median \(\mathrm{Me}=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_0} \times c\)

Here, \(\frac{N}{2}\) = \(\frac{50}{2}\) = 25

∴ the median lies in the class-boundary 35.5- 40.5 .

∴ x₁ = 35.5, F = 20 and f₀ = 15

c = Range of the class boundary consisting the median = 5

Putting these values in the above formula we get, \(\mathrm{Me}=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_0} \times c\)

∴ the required median = 37.17 marks.

Example 20. The frequency distribution of the monthly incomes of 300 workers of an industry is given below:

Determine

1. Arithmetic Mean,
2. Median and
3. Mode of the above frequency distribution

Solution:

The frequency distribution of the monthly incomes of 300 workers of an industry is given in the table

1. The mid-point of the class interval 140-150 is 145.

Let us take 145 as the origin (A) and class length (c) = 10 to be assumed as the unit.

Then, \(u=\frac{x-145}{10} \text { or, } \bar{x}=145+10 \bar{u}\)

Let us now draw the following table:

∴ \(\bar{u}=\frac{\sum u f}{\sum f}=\frac{-255}{300}=-0.85\)

\(\bar{x}\) = 145 + 10 x -0.85 = 145 -8.50= 136.50

 

or, Arithmetic mean = Rs. 136.50

2. By the formula, \(\mathrm{Me}=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_0} \times c\)

Here, \(\frac{N}{2}\) = \(\frac{300}{2}\) = 150

∴ the median lies in between 130 and 140, x₁= 1 30, F = 99, f₀ =100 and c = 10

∴ \(\mathrm{Me}=130+\frac{150-99}{100} \times 10=130+\frac{51 \times 10}{100}\) = 130 + 5.1 = 135.10

∴ the required median = Rs. 135.10

3. Since, the highest frequency = 100

∴ the mode lies in the class-interv al (130 – 140).

By the formula of Mode, \(M_0=x_1+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, \(x_1=130, f_0=100, f_{-1}=59, f_1=41 \text { and } c=10\)

∴ Mode = \(130+\frac{100-59}{200-59-41} \times 10=130+\frac{410}{100}=130+4 \cdot 10=134 \cdot 10\)

∴ the required mode = Rs. 134.10

Example 21. Find the mean, median, and mode of the following frequency distribution

Solution: Let us draw the following table

Let us take, A = 115 as the origin, which is the mid-value of the interval (110-120) and class length, c = 10

∴ u = \(\frac{x-115}{10} \Rightarrow \bar{x}=115+10 \bar{u}\)

∴ \(\bar{u}=\frac{\sum u f}{\sum f}=\frac{231}{300}=0.77\)

∴ \(\bar{x}\) = 150 + 10 \(\bar{u}\) = 1150 + 10 x 0.77 = 115 + 7.70 = 122.70

∴ Arithmetic mean =122.70

To determine the median, let us draw the following table:

Here, total frequency, n = 300 and \(\frac{n}{2}\) = \(\frac{300}{2}\) = 150

The number 150 lies between 50 and 150. .*. the median lies in the interval 110- 120.

Here, x_l = 110, F = 50,f₀ = 100 and c = 10

∴ Median, \(\mathrm{Me}=x_l+\frac{\frac{n}{2}-\mathrm{F}}{f_0} \times c=110+\frac{150-50}{100}\) x 10= 110 + 10 = 120.

∴ The required median = 120.

Here the mode lies in the interval (110- 120), since its frequency density is the highest.

The mode, \(\mathbf{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, \(x_l=110, f_0=100, f_{-1}=10, f_1=17 \text { and } c=10\)

∴ The mode =110 + \(\frac{100-10}{2 \times 100-10-17}\) x 10 = 110 + \(\frac{90 \times 10}{200-27}\)

= 110 + \(\frac{900}{173}=\frac{19030+900}{173}=\frac{19930}{173}\)

= 115.20 (approximated to two decimal points)

∴ The required mean = 122.70, median =120 and mode = 115.20.