Statistics

Mean:

It’s another name of ‘Average’. In exams, the word ‘arithmetic mean’ usually use instead of average or mean. So don’t confuse in these terms, these are different names of the same thing.

And we have already learned formula of average in Study Plan for Beginners. So here we’ll do it’s application.

Mean = ^Total⁄_Numbers

or

Mean = ^Sum⁄_Count

Let’s suppose a set S = {2, 4, 6, 8}

The average or mean of this set would be 5 by using the above formula. But there’s a short-cut if you see that there is a common difference between any two consecutive elements of the set or numbers. For instance in above case, the difference is common between any two consecutive elements (i.e 2). So in this case, always remember the short-cut to find mean of the set as follows:

Mean = ^{(First element + Last element)}⁄₂

In short, if you see that there is common difference between any two consecutive elements of a series of numbers or any set, always take average of first element and last element; this would give the same average.

For instance,

X = {2, 4, 6, 8, ….. 450}

What is the average of set X?

Solution:

As you see set starts with 2 and then proceed with subsequent addition of 2 for the next adjacent element and after element ‘8’ the doted marks shows this same trend apply until the series end to 450. So we need to just take the average of first and last element of this set which would give us the average of the set.

⇒ Mean of set X = ^{(2 + 450)}⁄₂

⇒                     = ^{(2 + 450)}⁄₂

⇒                     = ⁴⁵²⁄₂

⇒                     = 226 Answer

Important Note: Never use this short-cut when the difference between all two consecutive elements would not same. Also remember if difference between any two consecutive elements is same but except between one pair of consecutive element, again this short-cut wouldn’t applicable there.

For instance,

A = {2, 4, 6, 8, 12}

Here difference between all two consecutive elements is same except last two elements have difference which is different from other consecutive elements. So in this case, sort-cut is no more applicable.

Now, let’s learn how to find sum of a large evenly space set with the same short-cut.

Suppose the same set

X = {2, 4, 6, 8, ….. 450}

What is the sum of this set?

Solution:

First find mean of the set through the short-cut, which is 226 as we have already found.

Now, as we know

Mean = ^Sum⁄_Count

⇒ 226 = ^Sum⁄_Count ——————– (eq. 1)

We have value of Mean, and we need to find ‘Sum’, for that purpose we need to find ‘Count’ (i.e number of elements in set X).
For that purpose, always remember the formula bellow that many of you have studied in F.Sc Mathematics and that would also help you in ‘Sequences & Series’ topic later on.

a_n = a₁ + (n – 1)d

And remember that this formula is only useful where the difference between any two adjacent term/element is same.

Many of you don’t know this equation, so for those who didn’t know remember this formula and also it’s key variables as bellow:

a_n → n^th term of the sequence, series or set (i.e 10^th term, 20^th term, 100^th term or any term)

a₁ → first term of the series

n → number of elements of series from start till a_n (i.e 10, 20, 100 or any number)\

d → common difference between any two adjacent elements or terms.

Important point:

Very Important Note: Remember that in any set where the difference between any two adjacent terms is same, this is called evenly spaced set. Also remember that don’t confuse with the word ‘evenly spaced’ and avoid making mistake of perceiving this as only which has common difference of even number. Even if the common difference is odd number, it will be an evenly spaced set. So here evenly has almost meaning of equally.

Now, using the formula of n^th term, we can find the number of elements in this sequence. i.e as we know,

a_n = a₁ + (n – 1)d

⇒ 450 = 2 + (n – 1)2

⇒ 450 = 2 + 2n – 2

⇒ 450 = 2n

⇒ n = 225

Now, be putting this value of n in (eq. 1)

⇒ 226 = ^Sum⁄₂₂₅

⇒ Sum = 226 × 225

Note: Don’t worry about big multiplication. You will learn the short-cut techniques of multiplying big numbers like this in ‘Arithmetic’ topic later on. Here it’s mentioned only to understand the concept.

⇒ Sum = 50850 Answer

Average Range:

To understand this thing, let’s suppose 10 people, who have weights range from 110kg to 140kg, are participating in a wrestling competition. What is the minimum average weight of these 10 people? Also what is the maximum average weight of these 10 people?

Solution:

First, let’s solve for minimum average weight of the 10 wrestlers.

As we know,

Average = ^Sum⁄_Count ——————– (eq. 2)

To find minimum average weight, we need to make the sum of weight as minimum as possible. Because number of people (count) wouldn’t change and remain 10, so only what we can change is the weight. So you need to analyze with the weight range limit which is 110kg to 140kg. Right!

Now, for getting minimum sum of weight of all 10 people, you need to put 9 people weighing 110kg, while exactly 1 person weighing 140kg. You cannot put weight of all 10 people as 110, because of the weight range is 110-140, which means at least one person must have weight 110kg and at least one person must have weight 140kg, and remaining may have other weights that must be within the weight limit.

So, to find minimum possible sum of weights, we only have one way which is to put at least 1 person weighing 110kg and at least one person weighing 140kg, while remaining persons must weight least weight and that is 110kg.

Therefore, from (eq. 2)

⇒ Minimum Avg. weight of the 10 people = ^{(110 + 110 + 110 + 110 + 110 + 110 + 110 + 110 + 110 + 140)}⁄₁₀

In short form:

⇒ Minimum Avg. weight of the 10 people = ^{{110(9) + 140(1)}}⁄₁₀

⇒                                                           = ^{(990 + 140)}⁄₁₀

⇒                                                           = ⁽¹¹³⁰⁾⁄₁₀

⇒                                                           = 113kg Answer

Similarly, for maximum average weight, at least one person must weight 110kg and at least one person must weight 140kg, while remaining 8 persons must weight 140kg to make maximum sum of weights of 10 people.

⇒ Maximum Avg. weight of the 10 people = ^{{110(1) + 140(9)}}⁄₁₀

⇒                                                           = ^{(110 + 1260)}⁄₁₀

⇒                                                           = ⁽¹³⁷⁰⁾⁄₁₀

⇒                                                           = 137kg Answer

Aggregate Average:

This is very important concept and its questions come in such exams many times, even almost always!

As we know the word ‘aggregate’ means ‘total’. So when two or more averages are available for two or more variables, if we find the total final average of all averages available, this final average is known is aggregate average.

Let’s suppose in a GAT class the average score of 30 boys is 88, while that of 20 girls is 92. What is the average score of all the students in the class?

Solution:

As I said this is an aggregate average question scenario. Here you always need to follow the fix way as bellow:

From first variable boys:

As we know,

Boys average score= ^{Sum of boys score}⁄_{Number of boys}

⇒   88 = ^{Sum of boys score}⁄₃₀

⇒ Sum of boys score = 88 × 30

Now here’s a short way to multiply this

⇒ Sum of boys score = (80 + 8) × 30 = (80)30 + (8)30 = 2400 + 240

⇒ Sum of boys score = 2640 ———————— (eq. 3)

Now, From second variable girls:

As we know,

Girls average score = ^{Sum of girls score}⁄_{Number of girls}

⇒   92 = ^{Sum of girls score}⁄₂₀

⇒ Sum of boys score = 92 × 20

Again, here’s a short way to multiply this

⇒ Sum of boys score = 92 × (2 × 10) = (92 × 2) × 10 = 184 × 10

⇒ Sum of boys score = 1840 ———————— (eq. 4)

Now, we need to find the average score of all students, so

As we know,

Average score of all students = ^{Total score of all students}⁄_{Total number of students} ———————— (eq. 5)

Now, sum of (eq. 3) and (eq. 4) would give the total score of all students, while sum of number of boys and number of girls would give the total number of students

⇒ Total score of all students = sum of boys score + sum of girls score

⇒                           = 2640 + 1840 = 4480

Common equation:

Total number of students = number of boys + number of girls

⇒                           = 30 + 20 = 50

By putting these two values in (eq. 5), we’ll get

Average score of all students = ⁴⁴⁸⁰⁄₅₀ = ⁴⁴⁸⁄₅

⇒                           = 89.6 Answer

Important Point: Notice that simple average of (boys average score and girls average score) = ^{(88 + 92)}⁄₂ = 90

But through aggregate average, which gives the actual average score, it gives average 89.6 which is less than 90. The important point you need to understand here is that the difference in number of boys and number of girls has caused this fluctuation in aggregate average from simple average. And also notice that number of boys(whose average score is 88) are more than number of girls(whose average score is 92), so the aggregate average would tend to be more towards boys average score than girls average score. In other words, aggregate average would be less the simple average here, if you just analyze the situation. So through just analyzing, you can eliminate answer choices that are equals to or higher than 90. Some questions may take long calculations like the above, but there is always a short way by just analyzing situation and eliminate wrong answer choices.

Let’s do other scenarios of arithmetic mean.

Consider the same set S = {2, 4, 6, 8} with mean 5, as in start of this lecture,

If we add a new element x in set S, what should be the value of x such that the mean of the old set S would not change?

Solution:

After adding x to set S, we’ll get

S = {2, 4, 6, 8, x}

Now, as we know that, before adding x, the mean of set S was 5. Now imagine another set I = {5, 5, 5, 5} that also has mean 5. Now if we add another 5 to this set i.e I = {5, 5, 5, 5, 5}, it’s mean would still remain 5. So we learned that:

Rule 1: If we add same value equal to mean to any set, it’s mean would remain same.

Rule 2: If we add new element greater than mean of the previous set, it’s new mean would be greater than old mean (i.e new mean greater than mean before adding new element).

Rule 3: If we add new element smaller than mean of the previous set, it’s new mean would be less than old mean (i.e new mean less than mean before adding new element).

Let’s discuss another scenario (basically a word problem scenario) in which question from average comes in such exam.

Average of 5 different positive integer is 134, if one of the integer is discarded and 340 is added, the average becomes 150, what was the discarded integer?

Solution:

Always you must find the sum in such question as follows:

As we know,
Average of different integer = ^{Sum of these integers}⁄_{Number of integers}

⇒                           134 = ^{Sum of these integers}⁄₅

⇒         Sum of five integers = 134 × 5

Now, see how to calculate this without calculator

⇒ New Sum of five integers = (130 + 4)5           {By spliting 134 into easy muptiplyable form as (130 + 4)}

⇒                              = 650 + 20 = 670

Now, suppose x was the integer that was discarded

⇒ 670 – x + 350 = New sum of five different integers after some change

⇒ New sum after changes = 1020 – x ——————– (eq. 6)

Now, in question the new average of give different integers after the changes also give, so we just need to do the following this. And see how ways originate while you move on to next step and finally you will automatically reach to the answer, but it require practice to become expert in such things. So never discourage yourself, you are actually on the way to that level, and never disappoint even you couldn’t find the way by your own at this stage.

According to the given condition, it says after x is discarded and 340 is added, then new average becomes 140, so as we know,

Common equation:

New average after changes = ^{New sum after changes}⁄_{Number of integers}

By putting value of (eq. 6), we’ll get

⇒                   140 = ^{1020 – x}⁄₅

Now, you can solve it for value of x; With your increase in understanding and learning, I would reduce the calculation and jump directly to the answer so you may get opportunity to do the calculation by your own, because as I said in Study Plan for Beginners, Arithmetic is the key in whole quantitative section; if you wouldn’t improve this, you may never improve in quantitative section at all. So after solving it we’ll get

⇒ x = 320 Answer

Median:

Median means exactly at middle in place, position or rank. In other words, it’s a midpoint.

Consider a set, S = {2, 4, 6, 8, 10}

The median of this set is 6, because this comes at middle of all elements of this set.

But

If, S = {2, 4, 6, 8,}

Here, middle term doesn’t included in the set, rather there are two elements that come at middle; so in such a case where there are even number of elements in a set, always take the average of middle two terms. i.e in the given case,

Median = ^{(4 + 6)}⁄₂

⇒     = 5

Notice that the average of this set is also 5; this is because it’s an evenly spaced set.

Important point:

Important Note: Remember that in every evenly spaced set, mean and median would always equal. In other words, if we have one thing, we don’t need to waste our time to find other, because both are just equal so we’ll just have value of other because of the first.

Important point:

Important Note: Also remember that if the mean and median of a set are equal, then it doesn’t necessarily indicate that it’s an evenly spaced set. In short the reverse of previous ‘Important Note’ not necessary true. It’s just similar to a famous logical statement: “All mangoes are sweet, but all sweet are not necessarily mangoes”.

For instance, consider

X = {4, 4, 5, 5}

or

Y = {4, 4, 4, 5, 5, 5}

These above sets have mean and median equal, i.e 4.5, but these are not evenly spaced sets.

But if you take any evenly spaced set, you’ll see mean will always equals to median in that set.

Very Important Note: Remember that in order to find median of a set, always rearrange the set in ascending order (i.e from smaller element to larger element, while moving from left to right) if the set is not in ascending order by default.

Further in-depth scenarios of median like merging two or more sets and change in medians doesn’t come in exams like GAT, it usually come in exams like GMAT or GRE.

Mode:

Mode is simply the highest repeated element of a set or a highest repeated term of a series is known as mode of that set. In other words the element with highest frequency is mode of that set. In math term frequency means repetition. For instance if an element ‘x‘in a set comes 4 times, it’s frequency would be 4.

For instance,

A = {2, 4, 5, 5, 7, 8, 8, 8}

In set A, 5 comes twice, and 8 comes three times. Having the highest frequency, 8 would be the mode of set A.

Remember that if,

B = {2, 4, 5, 9, 15}

In set B, no element is repeating; here never say mode to be 0. As 0 is itself an element, i.e if 0 would repeat highest times in a set, then it’s mode would be 0 then. So here the mode of set B doesn’t exist.

Range:

It’s not hard and fast thing; it’s simply the difference between highest value and smallest value of a set or series. In other words range of any set cannot be negative. Minimum value of range is 0.

Let’s do a question of range:

Consider a set M = {x, x, x, y, y, y, x – y, 2x + 3y} where x < y. If the median of this set is 20, what is the range of this set?

Solution:

First see whether the set is arranged from leans to greatest, if not arrange it as follows:

As x < y, so x – y must be least element, so by arranging least to greatest we’ll get

M = {x – y, x, x, x, y, y, y, 3x + y}

Now, as there are even number of elements in the set, so median would be the average of middle two elements i.e,

Median = ^{(x + y}⁄₂

Also as it’s given that median is 20
So

⇒ ^{(x + y}⁄₂ = 20

⇒ x + y = 40 ——————- (eq. 7)

Now, as we now, range of set is maximum value – minimum value

⇒ Range = 3x + y – (x – y)               (never forget to place brackets after –ve sign)

⇒ Range = 3x + y – x + y           (As –ve sign multiplies when brackets open and change signs inside bracket)

⇒ Range = 2x + 2y = 2(x + y)

Now by putting value of (eq. 7) here, we’ll get

⇒ Range = 2(40) = 80 Answer

Standard Deviation:

Standard deviation means deviation from standard (i.e deviation from the mean). If a set is more spread from its mean, then its standard deviation would be higher; if a set is too much condense and spread not much far away from the mean, then its standard deviation would be less.

For instance,

Suppose A = {0.2, 0.4, 0.6, 0.8}

and B = {2, 4, 6, 8}

The SD (i.e Standard Deviation) of set A would be lower than that of set B, because each element of set A is differ from the mean by less than 1 unit. While each element of set B is differ from the mean by 1 and even greater than 1 for few elements.

By the way, you will never ask to find the SD of any set in such exams, rather you’ll ask to analyze whose SD is greater and by changes in elements (i.e addition or subtraction) what would be the impact on the standard deviation.

For general information the standard deviation is calculated as follows, but you never ask to calculate the standard deviation, rather you will ask to predict the change in already given standard deviation in the question.

Consider Set A = {2, 4, 6, 8}

As mean of this set is 5,

Standard Deviation = ^{Sum of the positive difference between each element and mean}⁄₄

Standard Deviation = ^{{|2 – 5| + |4 – 5| + |6 – 5| + |8 – 5|}}⁄₄

Here ‘|   |’ sign is called modulus (mod) which changes –ve sign to positive, but +ve sign remaing positive and 0 remain 0.

⇒ = ^{{3 + 1 + 1 + 3}}⁄₄ = ⁸⁄₄

⇒ = 2