avg

Contents

1. INTRODUCTION
2. MEASURES OF CENTRAL TENDENCY
   1. ARITHMETIC MEAN (AM)
   2. GEOMETRIC MEAN (GM)
   3. HARMONIC MEAN (HM)
   4. WEIGHTED AVERAGES
   5. MEDIAN
   6. MODE

Averages

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INTRODUCTION

Average is the most likely middle value of a data set. It is the mean value of a data set around which all the numbers are clustered and hence is a representative of the set. The average can be calculated by the sum of all values in the set divided by the total number of values.

If the number of observations is very large, looking at all of them might be very timeconsuming. An average thus gives us a better picture of a collection of data. For example, one of the parameters considered while selecting a good Bschool is the average package students manage to get. Prospective candidates may compare Bschools based on these average package figures. The bar graph shows a Bschool with an average package of Rs. 10 lakhs. It has been arrived at by adding all the salaries of the students and then dividing by the number of students.

Example 1:

A group of 5 friends scored 10, 12, 16, 20 and 18 in a class test. Find the average score. Solution: Hence, the average score is 15.2.

Example 2:

Consider a sequence of seven consecutive integers. The average of the first five integers is n. The average of all the seven integers is [CAT 2000] (1) n (2) n + 1 (3) K n, where K is a function of n (4) n + (2/7) Solution: Average of first five integers is n. The integers are (n – 2), (n – 1), n, (n +1), (n + 2). The last two numbers are (n + 3), (n + 4). The average of the seven numbers = (7n + 7)/7 = n + 1 Hence, option 2.

Example 3:

Rajiv is a student in a business school. After every test he calculates his cumulative average. QT and OB were his last two tests. 83 marks in QT increased his average by 2. 75 marks in OB further increased his average by 1. Reasoning is the next test, if he gets 51 in Reasoning, his average will be _____? [JMET 2009] (1) 63(2) 62(3) 61(4) 60(5) 59 Solution: Let the total marks of Rajiv and the number of tests he gave before giving QT be x and n respectively. 83 marks in QT increased his average by 2 75 marks in OB further increased his average by 1 He gets 51 in his next test Reasoning Solving equation (i) and (ii) we get the value of n = 10 and x = 610 From (iii) we get Average = 63 Hence, option 1.

REMEMBER:

• If the value of each item in a group is increased/decreased by the same value x, then the average of the group also increases/decreases by x. This concept is especially useful when dealing with ages. If the average age of a group of people is x years, then their average after n years will be (x + n) and their average age n years ago would have been (x – n) years. This is because with each passing year, each person’s age increases by 1.

• If the value of each item in a group is multiplied/divided by the same value x (where x ≠ 0 in the case of division), then the average of the group also gets multiplied/divided by x.

• The average of a group always lies between the smallest value and the largest value in the group.

• The net deficit due to the numbers below the average always equals the net surplus of the numbers above the average.

ALTERNATE METHOD TO CALCULATE AVERAGE

In most cases, we can avoid tedious calculations of adding all the numbers and dividing by the number of observations. The calculation of averages can be simplified using the assumed average method. Take some arbitrary average (A), which lies between the smallest and the largest value. Then, calculate the deviations (differences) of the given values from A and find the average of all these deviations. Add this to A to find the average.

Example 4:

Find the average of 102, 105, 92, 103, 96, and 98 using the assumed average method. Solution: The arbitrary number has to lie between 92 and 105. Let us take 100 as the arbitrary number. Let us now find the difference of each of the number from 100. 102 100 = 2 105 100 = 5 92 100 = 8 103 100 = 3 96 100 = 4 98 100 = 2 The sum of all deviations is 4. Hence, the average = 100 4/6 = 99.33 In this method, you end up adding smaller numbers and get the average conveniently.

EXPLORING AVERAGES WHEN NEW ITEMS ARE ADDED TO A SET

It is apparent that if all the values in a set are replaced with the average value, then the sum would remain unchanged.

For example, consider a set of 5 numbers; 22, 43, 51, 60 and 74. Their average is given by:

So, if we were to replace the 5 values in the set by 50, then the sum would still remain 250.

Now assume that a new item, whose value is 68, gets added to the above set. Hence, the new average will be (250 + 68)/6 = 318/6 = 53. However, there is a more convenient method of solving this.

We have already stated that, in the original set, the sum will not change if each of the 5 values were replaced by 50. Now the new number added is 68, which is (68 – 50) = 18 more than 50. This difference should now be equally distributed among 6 values, i.e. we add 18/6 = 3 to each of the original values. Hence, the new average will be 50 + 3 = 53.

i.e. New Average (after addition)

If the number added is less than the original average, then (Value of newly added item – Original Average) will be negative and the average will reduce.

Conversely,

Consider that we are told that the average of a set of 5 numbers is 50, and when another item is added to the set, the average increases by 3. You are asked to calculate the value of the newly added item. Here is how you should proceed:

The new average (of 6 numbers) is 50 + 3 = 53; i.e. the values could be 53, 53, 53, 53, 53 and 53. However, the old average was 50; so we assume that the values of all the numbers, except the newly added one, were (and still are) 50; i.e. the values are 50, 50, 50, 50, 50 and X, where X is the unknown value of the newly added item. So, the value of the newly added item will be (3 5) greater than 53; i.e. 53 + (3 5) = 68. Visually,

Example 5:

In a class, the average weight of students is 60 kg. If a student weighing 68 kg joins the class, the average weight increases by 1 kg. How many students were there in the class initially? Solution: The best way to solve this question is by assuming that each student weighs 60 kg. If now another student joins them and his weight is 68 kg, then he brings in extra 8 kg which gets distributed equally among 8 students since the new average increases by 1 kg. Hence, initially there were 7 students in the class.

Example 6:

The average weight of 10 oarsmen in a boat increases by 1 kg when one of the men who weighs 70 kg is replaced by a new man. What is the weight of the new man? Solution: The average weight of 10 men increases by 1 kg. Total weight increases by 10 1 = 10 kg Hence, the weight of the new man = 70 + 10 = 80 kg

Example 7:

The average weight of a school football team (consisting of 22 members including a goal keeper) decreases by half a kilogram if the goalie is not included. What is the goal keeper’s weight, if the average weight of the team initially was 60 kg? Solution: The average weight of the team including the goalie was 60 kg. This value can be obtained by assuming all 22 players to have a weight of 60 kg each. However, it is said that the average weight reduces by 0.5 kg when the goalie is not included; thus, we can assume that the remaining 21 players each weigh 60 – 0.5 = 59.5 kg. Hence, for the average of the whole team to be 60 kg, the goal keeper’s weight must be 60 + 21 0.5 = 70.5 kg. Hence, the goal keeper’s weight was 70.5 kg.

MEASURES OF CENTRAL TENDENCY

Measures of central tendency are used to measure the “central” or “expected” value of the data set.

Standard deviation and variance are called measures of spread. They tell us how spread the data is.

The five most common measures of central tendency are arithmetic mean, geometric mean, harmonic mean, median and mode.

ARITHMETIC MEAN (AM)

Arithmetic Mean is the representative value of a given set of values. It is the standard average, often called the “mean”. It is calculated by dividing the sum of all the values by the total number of values. It is used to find the central tendency of any random distribution of numbers.

Example 8:

The profits registered by a leading telecom operator for the past five years are Rs. 30 lakhs, Rs. 42 lakhs, Rs. 45 lakhs, Rs. 48 lakhs and Rs. 52 lakhs. Their closest rival had registered profits of Rs. 28 lakhs, Rs. 46 lakhs, Rs. 50 lakhs and Rs. 57 lakhs for the last four years. Which company registered better profits (in terms of average)? Solution: Let us use arithmetic mean, which is the average profit, to compare the profits registered by the two companies. The average profit of the rival company is higher. Hence, it has registered better profits.

REMEMBER:

• Arithmetic mean is not a very good measure of central tendency. This is because the mean is greatly influenced by outliers.

For example, one of the practical applications is the average score projected of a class test where one or two students scored very high marks.

If the scores of the students are 10, 12, 12, 12, 13 and 25 then the AM is 14 but 5 out of 6 scores are below this.

In such cases median should be used to represent the central value.

GEOMETRIC MEAN (GM)

The GM of n positive numbers is the mean calculated by taking the n^th root of the product of these numbers.

Example 9:

Find the geometric mean of the given set of numbers 3, 18, 12, 2. Solution: GM = (3 12 18 2)1/4 = (1296)1/4 = 6

If the numbers are multiplied then the geometric mean is used to compute the average. For example, geometric mean is used to calculate average growth rates or interest rates. The following example would make it clear.

Consider a scenario where a person deposits a certain amount of money in the bank. For four consecutive years, the bank compounds the principal by an interest rate of 10%, 12%, 14% and 15% respectively. Let us say, we have to calculate the net rate at which the bank compounded his principal.

Let the principal be Rs. 100. The amount at the end of the first year (denoted by, say, P₁) is given as,

P₁ = 100 1.1

Similarly, the values of P₂, P₃, and P₄ can be calculated and the principal amount at the end of the fourth year (P₄) is given as,

P₄ = 100 1.1 1.12 1.14 1.15

Now, let us consider the net rate of return as r%. Therefore, P₄ can be expressed as,

Comparing this with the earlier expression for P₄, we get the value of r as,

Also, if the population of a town increases by 10% one year and then decreases by 20% next year the average rate of increase/decrease per year will be the geometric mean of (100 + 10) = 110 and (100 20) = 80.

The average rate of decrease per year = 6.2%

We can check this value. If the population of the town initially is 100 then at the end of first year the population of the town will be 110.

In the second year the population of the town decreases by 20%.

The population of the town at the end of second year = 0.8 110 = 88

There has been a 12% decrease in 2 years, therefore, the average rate of decrease per year is approximately 6 %.

Example 10:

Ankush is planning to take a housing loan from a bank. He is confused about taking a floating or fixed rate of interest. The fixed rate of interest has remained constant at 12% for the last 3 years. The floating rate of interest was 12%, 15% and 14% in the last 3 years. Based on this data, should Ankush opt for floating or fixed rate of interest? Solution: The fixed rate of interest remained constant at 12% for the last three years. The floating rate of interest was 12%, 15% and 14% Geometric mean is used to calculate average interest rates. Average rate of interest for 3 years = (1.12 1.15 1.14)1/3 = 1.136 The average floating rate of interest for the three years is 13.6%. Hence, based on this data, Ankush should opt for fixed rate of interest.

HARMONIC MEAN (HM)

Harmonic mean of a set of numbers is given by the following formula:

For any two quantities, say a and b, HM is given by the formula:

Harmonic mean is the least of the three Pythagorean means (Arithmetic Mean, Geometric Mean and Harmonic Mean). This is because the Harmonic mean tends towards the smaller values of a given set of numbers. As opposed to the arithmetic mean, it is affected by small outliers.

In many situations involving rates and ratios the Harmonic Mean provides the best average.

For example, HM is generally used is to find the average speed when equal distances are covered at different speeds and the resultant resistance when two electric resistors are connected in parallel.

Explanation:

Suppose that a car travels a distance of D km; it travels the first half of the distance with a speed of a km/hr, and travels the second half with a speed of b km/hr.

Example 11:

The distance between city A and B is 300 km. Manoj covered the distance from city A to city B at the average speed of 60 km/hr. On the way back, he covered this distance at the average speed of 50 km/hr. What is the average speed for the entire journey of 600 km? Solution: Using harmonic mean for two numbers we get, = 54.54 km/hr

ALTERNATE METHOD TO CALCULATE AVERAGE SPEED

The average of two individual speeds (when the distances covered by the two speeds are equal) can also be calculated using the following method.

Consider the above example (i.e. Solved Example 11). Manoj travels from A to B at 60 km/hr (say a), and from B to A at 50 km/hr (say b).

Hence, the ratio of the two speeds (smaller speed : larger speed) will be 50 : 60 = 5 : 6 (= x : y, say)

Now, calculate the difference between the two speeds, |a – b| (i.e. 60 – 50 = 10) and divide that by the sum of the ratios, x + y (i.e. 5 + 6 = 11). Then multiply the result with x.

Hence, 10/11 5 = 4.54

This, when added to the smaller of the two speeds gives the average speed (i.e. 50 + 4.54 = 54.54 km/hr).

For most instances, this method should be less calculation intensive than the earlier methods.

Example 12:

Chandan plans a trip from Kanyakumari to Kashmir. He decides to travel half the distance by flight, and half of the remaining distance by train and the rest by a rented car. It is estimated that the aeroplane will move at a speed of 336 km/hr, the train at 60 km/hr and the car at 40 km/hr. What is Chandan’s estimated average speed during his trip? Solution: Notice that the distances travelled by the train and car are equal, and the combined distance covered by both train and car is equal to the distance covered by the aeroplane. To find the average speed of the train and car: Ratio of the smaller speed to the larger speed = 40 : 60 = 2 : 3 The difference between the two speeds = 60 – 40 = 20 Average speed of the train and car = 40 + [20/5 2] = 40 + 8 = 48 km/hr To find Chandan’s average speed for the trip: Ratio of the smaller speed to the larger speed = 48 : 336 = 1 : 7 The difference between the two speeds = 336 – 48 = 288 Average speed = 48 + [288/8 1] = 48 + 36 = 84 km/hr Hence, Chandan’s estimated average speed during the trip is 84 km/hr.

RELATION BETWEEN MEANS:

For two numbers a and b,

This means that GM is the Geometric Mean of AM and HM. Thus, AM, GM and HM are in geometric progression.

Now,

Hence, AM GM (AM = GM, when a and b are equal)

Since AM, GM and HM are in geometric progression and AM GM we can conclude that GM HM

AM GM HM

REMEMBER:

• The relationship between the Arithmetic Mean and the Geometric Mean (i.e. AM GM) is vitally important. Most problems based on inequalities can be solved using this relation.

WEIGHTED AVERAGES

The term ‘weight’ stands for the relative importance that is attached to the values. Sometimes the different values in the group have different ‘weights’ associated with them which is the relative importance of the values. For example, consider a situation where the average ages for different departments of an office are known and we have to calculate the combined average of the office. In this case, we would not take the simple average of the different average ages but rather we would use the following formula.

where S₁, S₂, ..., S_n are the number of employees in the departments which act as ‘weights’. The larger the group, the more its average will have an impact on the combined average.

The average in such a case is called the weighted average and is given by the following formula.

where w₁, w₂, w₃, … are the weights of the respective values.

REMEMBER:

• If all the weights are equal, then the weighted average is the same as simple average or arithmetic mean.

Example 13:

A test was given to divisions A, B, and C of grade 8. The average scores of division A, B, C are 25, 30 and 35 respectively. The strength of divisions A, B, and C is 20, 24 and 30 respectively. Find the average scores of grade 8. Solution: Since the strength of each division is different, we will use the weighted average to find the average scores of grade 8. Hence, the average score of grade 8 is 30.7.

Example 14:

Mumbai gets wheat from different regions. If they get 1200 kg from Nasik at the rate of Rs 30 per kg, 1500 kg from Ratnagiri at the rate of Rs 25 per kg and 2000 kg from Kolhapur at the rate of Rs 32 per kg, find the average cost of the wheat procured from the different regions. Solution: Here we will find the weighted average since the quantities procured from different regions are different. Weighted Average = 29.25 Hence, the average cost is Rs 29.25 per kg.

Example 15:

What is the average concentration of a mixture, if 3 litres of 26% sodium nitrate is added to 6 litres of 17% of sodium nitrate? Solution: We need to find the weighted average. Hence, the average concentration is

Example 16:

Three math classes; X, Y, and Z, take an algebra test. The average score in class X is 83. The average score in class Y is 76. The average score in class Z is 85. The average score of all students in classes X and Y together is 79. The average score of all students in classes Y and Z together is 81. What is the average for all three classes? [CAT 2001] (1) 81(2) 81.5 (3) 82(4) 84.5 Solution: Let there be x number of students in class X, y number of students in class Y and z number of students in class Z. 83x + 76y = 79(x + y) 4x = 3y Similarly, 76y + 85z = 81(y + z) 4z = 5y 20x = 15y = 12z x : y : z = 3 : 4 : 5 The average of all the three classes Hence, option 2.

Example 17:

A final year MBA student gets 50% in the exam and 80% in the assignments. If the exam should count for 70% of the final result and the assignment for 30%, what will be the final score of the student, if professors decide to use weighted harmonic mean to uneven performances? [FMS 2007] (1) 56.34%(2) 60.53% (3) 64.83%(4) 66.59% Solution: Let W1 and W2 be the weights assigned to the exam and assignment respectively. W1 = 70% and W2 = 30% Let X1 and X2 be percentage marks obtained by the student in the exam and assignment respectively. X1 = 50% and X2 = 80% Hence, option 1.

Example 18:

The endeavor of any responsible government is to keep a balance between the economic development and social development of the country. The economic development of a country can be measured in terms of GDP, whereas the social development of the country is measured in terms of Human Development Index (HDI). HDI is a simple average of three indices-life expectancy index, education index and GDP/SDP index. The education index is the combined index of two indices: Adult literacy rate with 2/3 weights. Combined gross enrollment rate with 1/3 weight. Consider the following data on India and China: What are the education index for China and the Human Development Index (HDI) for India? [FMS 2009] (1) Education index for China 94.13 and HDI for India 42.19 (2) Education index for China 94.63 and HDI for India 42.39 (3) Education index for China 91.13 and HDI for India 42.79 (4) Education index for China 91.63 and HDI for India 42.49 Solution: Education Index (EI) is the combined index of adult literacy rate and combined gross enrollment rate with weights of 2/3 and 1/3 respectively. We have been asked to find education index for China and HDI for India respectively. Since calculation for Education Index is simpler compared to HDI, we will first calculate the EI for China and eliminate answer options. For China, The only option having EI for China as 91.13 is option 3. Hence, option 3.

MEDIAN

Median is the middle value of a group of numbers arranged in an ascending or descending order.

If the number of values in a given set of data is even, then there will be two middle values say x and y.

REMEMBER:

• The median of a set of values may or may not be equal to, less than or greater than the mean of the set of values.

Example 19:

Find the median of the numbers 31, 43, 32, 45, 36, 42, 33, 39, 40. Solution: First arrange the numbers in ascending order or descending order. 31, 32, 33, 36, 39, 40, 42, 43 and 45 Hence, the median is the (9 + 1)/2 = 5th value, which is 39.

Example 20:

Find the median of the numbers 102, 99, 111, 101, 98, 87, 105, 100. Solution: First arrange the numbers in descending or ascending order. 111, 105, 102, 101, 100, 99, 98, 87 Since the number of observations is even, there are two middle values. The median is the average of the 4th and 5th value. Hence, the median is the average of 101 and 100, which is 100.5.

Example 21:

In the data set {2, 5, 7, 8, X}, the arithmetic mean is same as the median. Determine the value of X. Assume X 8. [JMET 2009] (1) 8(2) 10 (3) 13(4) 15 Solution: In the given data set {2, 5, 7, 8, X}, the median is same as the arithmetic mean of these number and furthermore the value of X 8. The median for the given data set would be 7. Now, X = 35 22 = 13 Hence, option 3.

MODE

Mode is the number that occurs most frequently in a given set of numbers. If two or more values appear the same number of times, then the data set does not have a unique mode. The mode of a set of values may or may not be equal to the mean or the median of the set of values.

Example 22:

Find the mode of the set of the numbers 1, 2, 3, 7, 5, 9, 2, 12, 2, 13, 4, 10, 2, 8, 2, 6? Solution: The numbers can be arranged in an ascending order as follows. 1, 2, 2, 2, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13 The number ‘2’ occurs maximum number of times. Hence, the mode is 2.