Posted on - 08-02-2017

Bank PO

“Average is a very simple but effective way of representing an entire group by a single value. .

“Average” of a group is defined as

Average =

“Sum of all the items in the group” means “sum of the values of all the items in the group”. .

Example, Let us say a cricketer played 9 innings in a year. Let us say he scored the following runs in those innings 35, 56, 124, 29, 0, 87, 98, 45 and 75. Then his average score (per innings) for the year is .

Total score/Number of innings = = 61

Similarly the average height of a class of students is equal to the sum of the heights of all the students of the class divided by the number of students in the class. .

Average is also called the “mean” or mean value of all the values. .

- If the value of each item is increased by the same value p, then the average of the group or items will also increase by p. .
- If the value of each item is decreased by the same value p, then the average of the group or items will also decrease by p. .
- If the value of each item is multiplied by the same value p, then the average of the group or items will also he multiplied by p. .
- If the value of each item is divided by the same value p (p > 0), then the average of the group or items will also be divided by p. .
- The average of a group of distinct items will always lie between the smallest value in the group and largest value in the group - i.e., the average will be greater than the smallest value and less than the largest value in the group. .

As already discussed, the average of a group
of items whose values are given can be found out by the rule given at the
beginning of this section. However, in most of the cases, we do not need to
perform such elaborate additions and divisions. The calculation of averages can
be simplified greatly by taking some arbitrary number (P) as a starting point,
take the deviations of the given items (Q_{1}) from this arbitrary
number, find the average of all these deviations (Q_{1} - P) and add it
to the arbitrary number (P) to give the correct average of the given items. .

If there are n items and they are denoted by
Q_{1}, Q_{2}, Q_{3}, ……Q_{n}, then the average
of these n items is given by

Average = P +

The extent to which this method will simplify the calculation will depend on the selection of the arbitrary value P. It should be selected in such a way that the positive and negative deviations cancel out each other to the extent possible. Then the final figure left for division will be relatively small making the division easier. .

For example, the cricketer that we considered above scored the following runs in seven innings: 35, 56, 45, 43, 67, 70 and 48. Now, to find his average, we take an arbitrary figure, say 50 and first find the deviations of each of the scores from this figure. The deviations of the scores from 50 are -15, +6, -5, -7, +17, +20 and -2. The sum of these deviations is +14. .

Hence the average of the cricketer’s scores is

50 + = 52

Please note that the number P (= 50 above) can be any value. Let us work out the same example taking a different value for P. Let us take P equal to 45. The deviations of the scores from P are -10, + 11, 0, -2, +22, +25 and +3. The sum of these deviations is 49. Hence the average is 45 + 49/7 = 45 + 7 = 52.

When two groups of items are combined together, then we can talk of the average of the entire group. However, if we know only the average of the two groups individually, we cannot find out the average of the combined group of items. .

For example, there are two sections A and B of a class where the average height of section A is 150 cm and that of section B is 160 cm. On the basis of this information alone, we cannot find the average of the entire class (of the two sections). As discussed earlier, the average height of the entire class is .

Since we do not have any information regarding the number of students in the two sections, we cannot find the average of the entire class. Now, suppose that we are given that there are 60 students in the section A and 40 students in the section B, then we can calculate the average height of the entire class which, in this case will be equal to = 154 cm.

This average height 154 cm of the entire class is called “weighted average” of the class. .

The above step in calculating the weighted average of the class can be rewritten as below:

=

3/5 and 2/5 are the weights multiplying the averages. Hence the average height of the entire class is called as weighted average. .

The monthly incomes of Ajay in January, February and March this year are Rs.3000, Rs.4000 and Rs.5000 respectively. Find his average monthly income for these three months. .

Average income =

=** **= Rs.4000

Three numbers have an average of 20. If two of the numbers are 14 and 28, the third number is .

Sum of the three numbers = (20) (3) = 60

Third number = 60 - (sum of the other two numbers) = 60 - 42 = 18

Vijay purchased 1 dozen mangoes at Rs.6 per dozen, 2 dozen mangoes of another variety at Rs.10 per dozen and 5 dozen mangoes of a third variety at Rs.6 per dozen. Find the average cost per dozen of mangoes purchased by Vijay.

Cost of 1 dozen mangoes = (1) (6) = Rs.6.

Cost of 2 dozen mangoes = (2) (10) = Rs.20 .

Cost of 5 dozen mangoes = (5) (6) = Rs.30.

Total cost of 8 dozen mangoes = Rs.56 .

Average cost per dozen of mangoes = = Rs.7

The average weight of a group of 4 girls is 25 kg. A girl joins them and the average weight of the group goes up by 1 kg. Find the weight of the girl who joined. .

Total weight of the 4 girls = (4) (25)

= 100 kg

Total weight of 5 girls after the girl joins them = (5) (26) = 130 kg

Weight of the girl who joined = 130 – 100

= 30 kg