WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency

WBBSE Class 10 Measures of Central Tendency Overview

In your previous classes, you have studied how different given data are summarised. Let the heights of 100 students of class X be taken. But to get a clear conception of this data, it is necessary to summarise it.

Firstly, we can prepare its frequency distribution. Secondly, to analyse the data, we can determine the different measures of the characteristics of the data.

We shall discuss here one of the characteristics of the given frequency distribution, which is a central tendency, i.e., in this chapter, we shall discuss different measures of central tendency.

You have already observed that the frequencies of any variable in the mid-positions Is higher than the frequencies of the variable in the end position.

WBBSE Solutions for Class 10 Maths

It will be seen that for the heights of the students, the number of students of middle heights is more than all. The number of very short-height or very long-height students is almost rare.

In a class, the number of very brilliant and of very dull students is always rare, most of the students are of middle memory.

In the case of incomes, it also found that the number of middle incomers is always high.

If we observe any frequency distribution keenly, we shall see that the frequencies of the first spell are low, then they gradually increase and become the highest at the mid-position and then it further gradually decreases along the end positions.

This is very characteristic of any given frequency distribution of any variable is applicable in almost all cases. Exception may arise for the mixture data.

This tendency of the frequency distributions towards the mid-positions is generally known as the central tendency.

To measure this central tendency we generally determine the average of the frequency distribution. The average is the real measure of any central tendency. You have some conceptions of different types of averages, like Arithipetic mean, Geometric mean, and Harmonic mean.

Amongst these means, we generally use the arithmetic mean as the measure of central tendency. The other two are generally used for special cases.

Except for the average, the other two which are generally used as the measures of central tendency are median and mode.

Understanding Mean, Median, and Mode

Arithmetic mean

By the term “average”, we generally mean the arithmetic mean. If the n-values of the variable x be x₁,x₂……….,x_n, then if their arithmetic mean be m_x or \(\bar{x}\) we write

\(\bar{x}=\frac{x_1+x_2+\cdots \cdots \cdots \cdots \cdots x_n}{n}=\frac{1}{n} \sum_{i=1}^n x_i \cdots \cdots \cdots(* 1)\)

i.e, the arithmetic mean of any variable is result of the division, when the sum of the values of the variable is divided by the number of the values.

Again, if the arithmetic mean is multiplied by the number of values we get the sum of the values of the variable, i.e.,

\(\bar{x} \times n=\sum_{i=1}^n x_i \cdots \cdots \cdots \cdots \cdots \cdots(* 2)\)

Again, we see that if the values of the variable be equal, then each value of the variable is equal to its arithmetic mean, i.e., \(\bar{x}\).

Thus \(\bar{x}\) is then the representative value.

Again since \(\bar{x}\) is the representative of the variable x, the frequency of x lies on \(\bar{x}\).

In this regard \(\bar{x}\) is said to be the measure of the location of x.

Statistics Chapter 1 Measures Of Central Tendency Examples

Finding Median in Even and Odd Data Sets

Example 1. Let a businessman have sold rice of 15 quintals, 19 quintals, and 20 quintals in three consecutive months. Then how many quintals of rice had he sold per month as an average?

Solution:

Given:

Let a businessman have sold rice of 15 quintals, 19 quintals, and 20 quintals in three consecutive months.

Here, let x is the variable of quantity of rice. Then the three values of x are x₁, x_2, and x₃ the unit being quintal each.

Then the arithmetic mean of these values of x is

\(\bar{x}=\frac{x_1+x_2+x_3}{3}=\frac{15+19+20}{3} \text { quintals }=\frac{54}{3} \text { quintals }=18 \text { quintals. }\)

Hence, he had sold 18 quintals of rice per month.

Example 2. Let the velocity of a bicycler in 10 consecutive hours are 17, 25, 30, 32, 28, 24, 20, 18, 16, and 10 in kilometers.

Solution:

Given:

Let the velocity of a bicycler in 10 consecutive hours are 17, 25, 30, 32, 28, 24, 20, 18, 16, and 10 in kilometers

Then, the average velocity,

\(\bar{x}=\frac{x_1+x_2+\cdots \cdots+x_{10}}{10}=\frac{17+25+30+\cdots \cdots \cdot+10}{10} \mathrm{~km}=\frac{220}{10} \mathrm{~km}=22 \mathrm{~km}\)

Hence, his average velocity of him = 22 km/h.

Now, if the person travels 10 hours with the same velocity, he would travel a distance of 22 x 10 km = 220 km in 10 hours.

Example 3. The marks obtained by 41 students in an examination in Math are given below:

Find the average of the marks obtained by them. Here, the total number of marks obtained

\(\sum_{i=1}^{41} x_i=15+25+32+\cdots \cdots \cdots+42=1,015\)

∴ the arithmetic mean, \(\bar{x}=\frac{\sum x_i}{n}=\frac{1,015}{41}\) = 24.8 (approx)

Now, for the convenient of computing of average of any variable we can transfer the origin or base of the variable x.

Thus, if we transfer the origin of x at A, then the new variable u = x- A

Here, \(\sum_{i=1}^n u_i=\sum_{i=1}^n x_i-n A \text { or, } \frac{1}{n} \sum u_i=\frac{1}{n} \sum x_i-A\)

or, \(\bar{u}=\bar{x}-A \text { or, } \bar{x}=\bar{u}+A \cdots \cdots \cdots \cdots \cdots \cdots(* 3)\)

Here, firstly the values of u_i should be determined by subtracting A from the values of x, then the value of \(\bar{u}\) is to be determined.

When A is added to \(\bar{u}\),we get the average \(\bar{x}\).

Example 4. The selling amount of money of a shopkeeper in 7 days of any week are given below (in rupees). 115,98, 102, 126,85,91, 107. Find the average of amount of rupees sold per day. Here, the values of the variable x, is near the value 100 of x.

Solution:

Given:

The selling amount of money of a shopkeeper in 7 days of any week are given below (in rupees). 115,98, 102, 126,85,91, 107.

If we transfer the origin A to 100, then the values of u are 15- 2, 2, 26, – 15,- 9, 7.

∴ \(\left.\bar{u}=\frac{1}{7}(15-2+2+26-15-9+7)=\frac{50-26}{7}=\frac{24}{7}=3.43 \text { (approx. }\right)\)

∴ \(\bar{x}\) = 100 + 3.43 = 103.43 (approx)

∴the required daily selling amount = Rs 103.43

Now, let x is a discrete variable. The possible values of x₁,x₂……….,x_n.

The frequency distribution of x are made from n values of x.

Let the frequencies of x₁,x₂……….,x_n be f₁,f₂……….,f_n, i.e., x₁ occurs f₁ times, x₂ occurs f₂ times,……. f_n occurs n-times and f₁,f₂……….,f_n = n.

Now, \(\sum_{i=1}^n f_i x_i=x_1 \times f_1+x_2 \times f_2+\ldots \ldots+x_n \times f_n\)

and = \(\mathrm{f}_1+\mathrm{f}_2+\ldots \ldots \ldots \ldots+\mathrm{f}_{\mathrm{n}}=\mathrm{n}\)

Arithmetic mean of \(x=\bar{x}=\frac{x_1 f_1+x_2 f_2+\cdots \cdots+x_n f_n}{f_1+f_2+\cdots \cdots+f_n}\)

= \(\frac{\sum_{i=1}^n x_i f_i}{\sum_{i=1}^n f_i}=\frac{1}{n} \sum_{i=1}^n x_i f_i \cdots \cdots \cdots \cdots \cdots \cdots(* 4)\)

This \(\bar{x}\) is the weighted mean of the variable x.

Example 5. Let a person bought 2 sharees of price Rs. 30 each, 3 sharees of price Rs. 35 each and 5 sharees of price Rs. 45 each. What is the average price of the sharees?

Solution:

Given:

Let a person bought 2 sharees of price Rs. 30 each, 3 sharees of price Rs. 35 each and 5 sharees of price Rs. 45 each.

Here, the total price of the sharees = Rs. (30 x 2 + 35 x 3 + 45 x 5)

= Rs. (60 + 105 + 225) = Rs. 390

and the total number, of sharees = 2 + 3 + 5 = 10.

∴ the average price of the sharees = Rs. \(\frac{390}{10}\) = Rs. 39.

Example 6. In a telephone exchange, the frequency distribution of the number of telephone calls per minute for any hour are given below:

Solution: Here, at first, the following table should be prepared:

Table: Determination of the number of telephone calls per minute.

∴ the average number of telephone calls \(\bar{x}\), where,

\(\bar{x}=\frac{\sum_{i=1}^9}{\sum_{i=1}^9 f i}=\frac{239}{60}=3 \cdot 98 \text { (approx.) }\)

Here, also the origin or base can be necessarily transferred.

Example 7. The average weight of the following frequency distribution is 117 kg.

Find the value of W.

Solution: Let us at first prepare the following table:

Average weight \((\bar{x})=\frac{\sum f x}{\sum f}=\frac{920+2 \mathrm{~W}}{10}\)

Here average weight = 117 kg(Given)

∴ \(117=\frac{920+2 W}{10}\)

or, 920 + 2W = 1170 or, 2W = 1170-920

or, 2W = 250 or, W = \(\frac{250}{2}\) or, W = 125 ∴ W = 125 kg

Example 8. In a shop there are 150 balls of 5 kinds of different radius. The frequency distribution of their diameters are as follows:

Find the average diameter of the balls.

Solution: Here, let 47 be the origin. Then, u = x – 47, where x = diameter in mm.

We at first prepare the following table:

Table: Determination of the average of diameters

∴ \(\frac{\sum u f}{\sum f}=\frac{17}{150}=0 \cdot 11 \text { (approx.) }\)

∴If the average of diameters be \((\bar{x})\), then \((\bar{x})\) = \((\bar{u})\) +47 = (0.11 + 47) mm = 47.11 mm (approx.)

In the case of continuous variable, if we want to find the average from the frequency distribution, it is quite impossible to find the correct value of the average.

Because, here we know the frequency of a certain class interval, but we do not know the individual values of each and every.

Here, we have to assume that all the values of the class interval are centered in the mid-point of that class, i.e… we should assume the frequency of the mid-point as the frequency of that class interval involved.

Thus, if 25-29 be any class interval and 10 be its frequency, then we should assume that the frequency of the mid-value 27 of the class interval 25-29 is 10.

By this assumption, we can find the approximate value of \((\bar{x})\).

Moreover, if we take the mid-value of any class interval in the middle position as the origin, the calculation becomes easier.

If the class length of the frequency distribution be equal, i.e., if the ranges of all the class intervals are equal, then it will be more convenient when the class lengths are taken as the unit of measurement.

Thus, if A be the origin and class interval c be the unit, then the new variable u will be:

u= \(\frac{x-A}{c}\) or, \(x=\mathrm{A}+c u \text { or, } x . f=A . f+c . u_f f . \text { or, } \sum x f=A \sum f+c \sum u f\)

or, \(\frac{\sum x f}{\sum f}=A+c \frac{\sum u f}{\sum f} \text { or, } \bar{x}=A+c \cdot \bar{u} \cdots \cdots \cdots \cdots \cdots \cdots(* 5)\)

Examples of Mode Calculation

Example 9. The frequency distribution of the heights (in cm) of 100 students of class X in a school. Find the average height of the students.

Solution:

Given

The frequency distribution of the heights (in cm) of 100 students of class X in a school.

Here, the class boundaries are 139.5- 144.5, 144.5- 149.5,…….179.5- 184.5.

The class mid-values 142, 147…….182 and class-length = 5.

The mid-value 162 of the class interval 160-164 is assumed to be the origin and 5, the class-length is assumed to be the unit, then \(u=\frac{x-162}{5} \text { and } \bar{x}=162+5 \bar{u}\)

Let us now prepare the following Table:

Table: Determination of the average heights of 100 students from frequency distribution:

∴ \( u=\frac{\sum u \cdot f}{\sum f}=\frac{8}{100}=0.08\)

∴ \(\bar{x}=162+5 . \bar{u}\) = 162 + 5 x 0.08 = 162 + 0.40 = 162.40 cm

∴ The average height, of 100 students is 162.4 cm.

Example 10. The frequency distribution of the marks obtained by 76 students in a math exam. Find the average of marks obtained.

Solution:

Table: Determination of average marks obtained in math exam of 76 students

∴ \(u=\frac{12}{7.6}=0 \cdot 158\)

∴ \(\bar{x}\) = 25.5 + 10 x 0. 158 = 25.5 + 1.58 = 27.08 (approx)

Properties of Arithmetic mean

1. The sum of the differences of the values of the variable from the arithmetic mean is zero.

Let the n values of x are x₁,x₂……….,x_n.

The arithmetic mean of them is \(\bar{x}\) where \(\bar{x}=\frac{\sum x_i}{n}\)

The deviation of n-values of x from \(\bar{x}\) are \(x_1-\bar{x}, x_2-\bar{x}, \ldots \ldots \ldots x_n-\bar{x}\).

∴ \(\sum\left(x_i-\bar{x}\right)=\left(x_1-\bar{x}\right)+\left(x_2-\bar{x}\right)+\ldots \ldots+\left(x_n-\bar{x}\right)\)

= \(\left(x_1+x_2+\ldots \ldots \ldots \ldots+x_n\right)-n \bar{x}=n \bar{x}-n \bar{x}=0\)

Applications of Central Tendency in Real Life

Example 11. \(\bar{x}\) = 18 quintal.

Here, the deviations are 15- 18, 19 -18, 20- 18,

i.e.,- 3, 1, and 2. Then, the sum of- 3, 1, and 2 is zero.

Example 12.\(\bar{x}\) = 22 km.

Their deviations are – 5, 3, 8, 10, 6, 2,- 2,- 4,- 6,- 12, the sum of whose is zero.

2. If the arithmetic mean of the variable x be \(\bar{x}\) and y and x are linearly related by y = a + bx, then \(\bar{y}\) = a+bx.

Here, y_i = a + bx_i, i = 1, 2,…….n, then

\(\sum y_i=n a+b \sum x_i, \text { or, } \bar{y}=a+b \bar{x} \text {. }\)

Example 13. The arithmetic mean of the temperatures in the centigrade scale of seven different days is 32.5°C. What will be the arithmetic mean in Fahrenheit scale?

Solution:

Given:

The arithmetic mean of the temperatures in the centigrade scale of seven different days is 32.5°C.

Let the temperatures of seven days be c₁,c₂……….,c_n in °C.

As per question, \(\overline{\mathrm{C}}=\frac{\sum \mathrm{C}_i}{7}\) = 32.5

We know that in Fahrenheit scale if the temperature be F, then \(F=32+\frac{9}{5} \bar{C}\)

∴ \(F=32+\frac{9}{5} \bar{C}\) = 32 + \(\frac{9}{2}\) x 32.5 = 32 + 9 x 6.5 = 32 + 58.5 = 90.5

∴ In the Fahrenheit scale, the average is 90-5°F.

3. If the arithmetic mean of two variables x and y be \(\bar{x}\) and \(\bar{y}\) respectively and if z =ax + by, then
\(\bar{z}\) = a\(\bar{x}\) + b \(\bar{x}\).

We have z = ax + by

∴ \(\mathrm{z}_{\mathrm{i}}=\mathrm{ax}_{\mathrm{i}}+\mathrm{by}_{\mathrm{i}}, \mathrm{i}=1,2, \ldots, \mathrm{n}, then \sum z_i=a \sum x_i+b \sum y_i\)

or, \(\frac{1}{n} \sum z_i=a \cdot \frac{1}{n} \sum x_i+b \cdot \frac{1}{n} \sum y_i\)

or, \(\bar{z}=a \bar{x}+b \bar{y}\) [because \(\frac{1}{n} \sum z_i=\bar{z}\), \(\frac{1}{n} \sum x_i=\bar{x}\) and \(\frac{1}{n} \sum y_i=\bar{y}\)

Example 14. The arithmetic means of the marks obtained by the students of a school in English and Bengali are 40.25 and 48.50 respectively. Find the averages of the sums of the marks obtained in English and Bengali.

Solution:

Given:

The arithmetic means of the marks obtained by the students of a school in English and Bengali are 40.25 and 48.50 respectively.

Here, if x be the sum of the marks obtained in English and y be that in Bengali, then their sum z will be x + y.

∴ \(\bar{z}\) = a \(\bar{x}\) + b \(\bar{y}\) = 40.25 + 48.50 = 88.75

∴ the average of the sums is 88.75.

4. For the variable x, if the arithmetic mean of two group of n₁, and n₂ numbers of values of \(\bar{x}_1\), and \(\bar{x}_2\) respectively, then their arithmetic mean of (n₁ + n₂) number of values is given by \(\bar{x}\) = \(\bar{x}=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\)

Here, let the n₁ number of values of x be x₁₁, x₁₂,…………x₁n₁ and let the n₂ number of values of x be x₂₁,x₂₂,………,x₂n₂.

Now, if the arithmetic means of two groups be \(\bar{x}_1\) and \(\bar{x}_2\), then \(\bar{x}_1\) = \(\frac{x_{11}+x_{12}+\cdots \cdots+x_{1 n_1}}{n_1}\) and \(\bar{x}_2=\frac{x_{21}+x_{22}+\cdots \cdots+x_{2 n_2}}{n_2}\)

Also, if the collective average of them be \(\bar{x}\) then

\(\bar{x}=\frac{\left(x_{11}+x_{12}+\cdots \cdots+x_{1 n_1}\right)+\left(x_{21}+x_{22}+\cdots \cdots+\dot{x}_{2 n_2}\right)}{n_1+n_2}=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\)

Example 15. There are 25 boy students and 20 girl students in the school. The average of mark in math, of 25 boy students is 52 and the average of marks in math of 20 girl-students is 48. Find the average of marks obtained by the students of the school as a whole.

Solution:

Given:

There are 25 boy students and 20 girl students in the school. The average of mark in math, of 25 boy students is 52 and the average of marks in math of 20 girl-students is 48.

Here n₁= 25 and n₂ = 20

\(\bar{x}_1\)= 52 and \(\bar{x}_2\) = 48.

∴ the collective average

\(\bar{x}=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}=\frac{25 \times 52+20 \times 48}{25+20}\)

= \(\frac{1300+960}{45}=\frac{2260}{45}=50 \cdot 2 \text { (approx.) }\)

Geometric Mean

Let \(x_1, x_2, \ldots, x_n\) be n values of x.

Then if the geometric mean be \(x_g=\left(x_1 \cdot x_2 \cdot \cdots \cdots \cdots x_n\right)^{\frac{1}{n}} \cdots \cdots \cdots \cdots \cdots \cdots(* 6)\)

In order to find the average of the proportion of more than one of two variables, the geometric mean is widely used.

Let \(x_1, x_2, \ldots, x_n\) be n values of x and \(y_1, y_2, \ldots, y_n\) be n values of y.

Now, taking the proportion of x and y, let us prepare a new variable z, where z = \(\frac{x}{y}\)

The geometric mean of the variable z, is

\(z_g=\left(z_1 \cdot z_2 \ldots \ldots \ldots z_n\right)^{\frac{1}{n}}=\left(\frac{x_1}{y_1} \cdot \frac{x_2}{y_2} \ldots \ldots \frac{x_n}{y_n}\right)^{\frac{1}{n}}=\frac{\left(x_1 x_2 \ldots \ldots \ldots x_n\right)^{\frac{1}{n}}}{\left(y_1 y_2 \ldots \ldots \ldots y_n\right)^{\frac{1}{n}}}=\frac{x_g}{y_g}\)

To find the average of compound interests, this geometric mean is widely used

Example 16. Find the geometric mean of the following numbers: 18, 40, 25, and 45. We have a total number of numbers is 4.

Solution: ∴Geometric mean

= \((18 \times 40 \times 25 \times 45)^{\frac{1}{4}}=\{(3 \times 3 \times 2) \times(2 \times 2 \times 2 \times 5) \times(5 \times 5) \times(5 \times 3 \times 3)\}^{\frac{1}{4}}\)

= \(\left(3^4 \times 2^4 \times 5^4\right)^{\frac{1}{4}}=3 \times 2 \times 5=30\)

Harmonic mean

Let \(x_1, x_2, \ldots, x_n\) be the n-values of x are given.

If the harmonic mean of them be x_n, then

\(x_n=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots \cdots+\frac{1}{x_n}}=\frac{n}{\sum_{i=1}^n \frac{1}{x_i}} \cdots \cdots \cdots \cdots \cdots \cdots(* 7)\)

Sometimes, the given variable is given as the rate of some objects.

For example, velocity per hour, price per kilogram, weight per cubic foot, etc.

In these cases, the harmonic mean is used.

Example 17. A train travels the first 30 miles at a speed of 15 miles/hour and the next 30 miles at a speed of 30 miles/hour. Find the average velocity of the train.

Solution:

Given:

A train travels the first 30 miles at a speed of 15 miles/hour and the next 30 miles at a speed of 30 miles/hour.

Here, the total distance = (30 + 30) miles = 60 miles.

Total times required = \(\left(\frac{30}{15}+\frac{30}{30}\right)\) hours =(2 + 1) hours = 3 hours.

∴ the average velocity of the train = \(\frac{60}{3}\) miles / hour = 20 miles / hour.

By using the formula of harmonic mean, we can also find the required average.

The train has travelled same distance with two types of velocities.

These two velocities are 15 miles/hour and 30. miles/hour. The harmonic mean of these two types of Velocities

= \(\frac{2}{\frac{1}{15}+\frac{1}{30}}=\frac{2}{\frac{2+1}{30}}=2 \times \frac{30}{3}=20\)

∴ the required average = 20 miles/hour

Example 18. A shopkeeper bought rice of Rs. 30 at a rate of Rs 1.5 per kg and of Rs 30 at a rate of Rs. 4 per kg. Then what is the value of mixed rice per kg?

Solution:

Given:

A shopkeeper bought rice of Rs. 30 at a rate of Rs 1.5 per kg and of Rs 30 at a rate of Rs. 4 per kg.

Here, the total amount of buying price of the shopkeeper = Rs. (30 + 30) = Rs. 60.

The quantity of the total rice bought

= \(\frac{30}{1 \cdot 5}+\frac{30}{4} \mathrm{~kg}=20+7 \frac{1}{2} \mathrm{~kg}=27 \frac{1}{2} \mathrm{~kg}\)

∴ the price of the mixed rice per kg = Rs \(\frac{60}{27 \frac{1}{2}}\)

= Rs. \(\frac{60}{\frac{55}{2}}\) = Rs \(\frac{24}{11}\) = Rs 2.18 (approx.)

∴ The required rate of price = Rs. 2.18 per kg.

We can also find the result by using the harmonic mean.

According to the formula, the H.M.

= \(\frac{2}{\frac{1}{1 \cdot 5}+\frac{1}{4}}=\frac{2}{\frac{1}{1 \frac{1}{2}}+\frac{1}{4}}=\frac{2}{\frac{1}{\frac{3}{2}}+\frac{1}{4}}=\frac{2}{\frac{2}{3}+\frac{1}{4}}\)

= \(\frac{2}{\frac{8+3}{12}}=\frac{2 \times 12}{11}=\frac{24}{11}=2 \cdot 18\)

∴ the average price of rice = Rs. 2.18 per. kg

Example 19. The frequency distribution of the monthly salaries of 80 workers of a company are given below. Find the arithmetic mean of the salaries of the workers.

Solution:

Given

The frequency distribution of the monthly salaries of 80 workers of a company are given in the table.

Here, we have to prepare class boundaries and class mid-values.

Due to the different lengths of the class intervals, by changing the origin and order, it is impossible to find out the average.

So, we have to prepare the following table at first:

Determination of average of the salaries of 80 workers from the given frequency distribution

∴ \(\bar{x}=\frac{\sum x f}{\sum f}=\frac{32390}{80}=404 \cdot 87\)

∴ the average of monthly salary of 80 workers = Rs. 404.87

Example 20. There are two groups A and B in a shop. In group A, 50 toys have been sold the average price of each toy being Rs. 15. In group B there also have been sold 30 toys, the average price of which is Rs. 20 each. Find the average of the average selling price of the total 80 toys sold in the shop.

Solution:

Given:

There are two groups A and B in a shop. In group A, 50 toys have been sold the average price of each toy being Rs. 15. In group B there also have been sold 30 toys, the average price of which is Rs. 20 each.

Here, the total selling price of 50 toys in group A = Rs. 50 x 15 = Rs. 750

The total selling price of 30 toys in group B = Rs. 30 x 20 = Rs. 600

∴ The total selling price = Rs. (750 + 600) = Rs. 1350

∴ the required average = Rs. \(\frac{1350}{50+30}\) = Rs.\(\frac{1350}{80}\) = Rs. 16.87 (approx.)

Example 21. The frequency distribution of the degree of the drugs of 110 patients to be cured are given below Here the number of patients being cured by taking 20 tablets and the number of patients being cured by taking 32 tablets are unknown. But it is known that the number of tablets required to be cured per patient is 20 tablets. Find the two unknown frequencies.

Solution:

Given

The frequency distribution of the degree of the drugs of 110 patients to be cured are given below Here the number of patients being cured by taking 20 tablets and the number of patients being cured by taking 32 tablets are unknown. But it is known that the number of tablets required to be cured per patient is 20 tablets.

Let the unknown frequencies be f₁ and f₂.

Also, let the origin of the variable (here tablets) be 20 and the unit of degree be 4, the class length, then

\(u=\frac{x-20}{4} \text { and } x=20+4 \bar{u}\)

To find the average, we prepare the following table

The total frequency = 110, ∴ 110 = 90 + f₁ + f₂ or, f₁+ f₂ = 20…….(1)

Also, the number of average tablets = 20

∴ 20= 20 + 4 . \(\bar{u}\) or, 0 = 4. \(\frac{\left(-42+3 f_2\right)}{110}\)

or,- 42 + 3f₂ = 0 or, 3f₂ = 42 or, f₂ = \(\frac{42}{3}\) = 14

∴ from (1), we get, f₁ + 14 = 20 or, = 20- 14 = 6

∴ the unknown two frequencies are 6 and 14.

Visual Representation of Mean, Median, and Mode

Example 22. The frequency distributions of the weekly wages of some laborers of industry are given below: The number of laborers having Rs 25 weekly wages is unknown. But it is known that the average weekly wages of a labourer is Rs 27.50. Find the unknown frequency.

Solution: Here, assuming 30 as the origin and the class length 5 as the unit of degree, we get, the new variable u,

where, \(u=\frac{x-30}{5}\), i.e., x=30+5 \(\bar{u}\)

Let the unknown frequency be f. Then we prepare the following table

∴ \(\bar{u}=\frac{\sum u \times f}{\sum f}=\frac{-17-f}{52+f} \text { and } \bar{x}=30+5 \cdot \frac{-17-f}{52+f}\)

or, 27.5 = 30.5 . \(\frac{-17-f}{52+f}\) [x = Rs. 27.5]

or,-0.5 = \(\frac{-17-f}{52+f}\) or, 0.5(52 + f) = 17 + f, or, 26 + 0.5 f = 17 + f.

or, 26-17 = f – 0.5f, or, 9 = 0.5f or, f = 18, which is the required frequency.

Example 23. In a business house, the weekly salaries of 30 employees are given below:

The rate of bonus of that house are given in the following table :

Find the arithmetic mean of. the bonus of the employees.

Solution: We have to prepare, firstly the frequency distribution table.

In this table, the class intervals are 61-75, 76-90,91-105, etc.

The frequencies of these intervals are f₁, f₂, f₃,……..etc. and the rate of bonus etc. are y₁, y₂, y₃, … etc.

Then the arithmetic mean of bonus, \(\bar{y}=\frac{\sum y_i f_i}{\sum f}\)

To find the arithmetic mean, let us prepare the frequency table:

In the case of the variable, let us assume 1 00 as the origin and class length 254 as the unit then \(u=\frac{y-100}{25}\) and \(\bar{y}=100+25 \cdot \bar{u}\).

Here, \(\bar{u}=\frac{\sum u \cdot f}{\sum f}=\frac{24}{30}=\frac{4}{5}\)

∴ \(\bar{y}\) = 25. \(\frac{4}{5}\) = Rs. (100 +20) = Rs. 120

∴ the average of bonus is Rs. 120

Example 24. There are 50 workers in a small industry. The daily income of them is Rs. 9. Amongst them, the daily income of the 30 workers, working in the morning is Rs. 8. Then find the daily average income of the workers working in the evening.

Solution:

Given:

There are 50 workers in a small industry. The daily income of them is Rs. 9. Amongst them, the daily income of the 30 workers, working in the morning is Rs. 8.

Here, the total number of workers, n = 50

The number of workers in the morning n₁ = 30

∴ The numbers of workers in the evening n₂ = n- n₁ = 50- 30 = 20

The average daily of the total workers = Rs. 9

The average daily income of the workers in the morning, \(\bar{x}_1\) = Rs. 8

According to the law of collective average,

\(\bar{x}=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\)

∴ \(9=\frac{30 \times 8+20 \bar{x}_2}{30+20} \text { or, } 9=\frac{240+20 \bar{x}_2}{50}\)

or, 450 = 240 + 20 \(\bar{x}_2\) or, 20 \(\bar{x}_2\) = 450-240

or, 20 \(\bar{x}_2\) =210 or, 20 \(\bar{x}_2\) = 210 or, \(\bar{x}_2\) = \(\frac{210}{20}\) = 10.50

∴ The daily income of the workers in the evening = Rs. 10.50.

Example 25. Find the day of the average absence of each and every student from the following given table:

Solution: Here, the upper boundaries of the class intervals and less-than-type cumulative frequencies are given.

First, the following table is to be prepared:

∴ \(\bar{u}=\frac{\sum u . f}{\sum f}=\frac{-40}{630}=-\frac{4}{63}=-0.0635 \text { (approx.) }\)

∴ \(\bar{x}\) = 12 + 5 \(\bar{u}\) = 12 + 5(-0.0635 = 12-0.3175 = 11.6825

∴ The average absence of the students = 11.68 days.

Example 26. The arithmetic mean of 200 values of a variable is 50. Later on, it is found that two values 92 and 8 have been taken wrongly instead of 192 and 88 respectively. After correction, find the actual arithmetic mean of the given values.

Solution:

Given:

The arithmetic mean of 200 values of a variable is 50. Later on, it is found that two values 92 and 8 have been taken wrongly instead of 192 and 88 respectively.

Here, the summation of the values of the variable = \(=\sum_{i=1}^{200} x_i=200 \bar{x}\) = 200 x 50 = 10000

Taking correct values, the summation = 10000-92-8+ 192 + 88= 10180

∴  the corrected arithmetic mean = \(\frac{10180}{200}\) = 50.9

Example 27. The average of the weights of 150 students is 50 kilogram. The average of the weights of the boy students is 55 kilograms and the average of the weights of the girl students is 42.5 kilogram. Find the number of boy and girl students.

Solution:

Given:

The average of the weights of 150 students is 50 kilogram. The average of the weights of the boy students is 55 kilograms and the average of the weights of the girl students is 42.5 kilogram.

Here, the average of the weights of the total students is \(\bar{x}\) = 50 kilogram

Total number of students n = 150

The average of weights of the boy students, \(\bar{x}_1\) = 55 kilogram

The average of weights of the girl students, \(\bar{x}_2\)=42.5 kilogram

We have to find the number of boy students, and the number of girl students.

It is known that n = n₁ + n2 or, 150 = n₁ + n₂……..(1)

Putting these values in the formula of collective arithmetic mean, we get

\(50=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2} \text { or, } 50=\frac{n_1 \times 55+n_2 \times 42.5}{150}\)

or, 7500 = 55n₁ + 42.5n₂………(2)

Solving (1) and (2), we get n₁ = 90 and n₂ = 60

∴ The number of boy students = 90 and the number.of girl students = 60

Example 28. If a and b be two positive numbers, then prove that A.M ≥ G.M. ≥ H.M. where

A.M. = Arithmetic mean
G.M. = Geometric mean
H.M = Harmonic mean

Solution: If a and b be two positive numbers, then

A.M. = \(\frac{a+b}{2}, \mathrm{G} \cdot \mathrm{M} .=(a b)^{\frac{1}{2}}=\sqrt{a b}\) and

H.M = \(\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2}{\frac{a+b}{a b}}=\frac{2 a b}{a+b}\)

Now, A.M.-G.M.= \(\frac{a+b}{2}-\sqrt{a b}=\frac{1}{2}(a+b-2 \sqrt{a b})=\frac{1}{2}(\sqrt{a}-\sqrt{b})^2\)

We know (√a- √b) may be positive or negative. But its square is always positive.

The value of (√a- √b) is 0 when √a = √b.

∴ \(\frac{1}{2}(\sqrt{a}-\sqrt{b})^2 \geq 0\)

or, A.M.- G.M ≥ 0. or. A.M > G.M……(1)

Substituting \(\frac{1}{a}\) and \(\frac{1}{a}\) instead of a and b respectively and the A.M. > G.M

∴ \(\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right) \geq\left(\frac{1}{a} \cdot \frac{1}{b}\right)^{\frac{1}{2}}\)

or, \(\frac{a+b}{2 a b} \geq \frac{1}{\sqrt{a b}}\),

or, \(\sqrt{a b} \geq \frac{2 a b}{a+b}\)

∴ G.M. ≥ H.M,………(2)

From the above relations (1) and (2). we get. A.M. ≥ G.M. ≥ H.M.

∴ A.M.≥ G.M ≥ H.M (Proved)

Example 29. The AM of two positive numbers is 25 and their GM is 15. Find the two numbers.

Solution:

Given:

The AM of two positive numbers is 25 and their GM is 15.

Let the two positive numbers be a and b

∴ their AM = \(\mathrm{AM}=\frac{a+b}{2}\) and GM = \((a b)^{\frac{1}{2}}=\sqrt{a b}\)

As per question, \(\mathrm{AM}=\frac{a+b}{2}\) or, a + b = 50………(1)

and =15 or, ab = 225……..(2)

From (1) we get, a = 50- b

∴ from (2) we get (50- b) b = 225, or, 50b- b² = 225

or, b²– 50b+ 225 = 0 or, b²– 5b- 45b + 225 = 0

or, b (b- 5)- 45 (b- 5) = 0 or, (b- 5) (b- 45) = 0

∴ either b-5 = 0or, b = 5 or, b- 45 = 0 or, b = 45

∴ b = 5 and b = 45

∴ the required numbers are 5 and 45.