# Arithmetic Mean for Ungrouped Data – Statistics

MEASURING THE CENTRE DATA

We can get a better understanding of a data set if we can locate the middle or centre of the data, and also get an indication of its spread or dispersion. Knowing one of these without the other is often of little use.

There are three statistics that are used to measure the centre of a data set. These are the mode, the mean, and the median.

The Mean

Whenever most people talk about an average, this is the one they… mean!
How to work out the Mean:
2. Divide this total by the number of data values

The mean of a data set is the statistical name for the arithmetic average.

The mean of the data is the average if you add all the values and divide by the number of values. We use the symbol for the mean.

The mean gives us a single number which indicates a centre of the data set. It is usually not a member of the data set.

For example, a mean test mark of 73% tells us that there are several marks below 73% and several above it. 73% is at the centre, but it is not always the case that one of the students scored 73%.

Suppose x is a numerical variable. We let:
xi be the i-th data value,
n be the number of data values in the sample or population
represent the mean of a sample, so

μ represent the mean of a population, so

In many cases we do not have data from all of the members of a population, so the exact value of μ is often unknown.
Instead we collect data from a sample of the population, and use the mean of the sample as an approximation for μ.

Definition: Mean
The mean is the sum of a set of values, divided by the number of values in the set. The notation for the mean of a set of values is a horizontal bar over the variable used to represent the set, for example 92. The formula for the mean of a data set {x1;x2;…;xn } is:

The mean is sometimes also called the average or the arithmetic mean.

Worked example: Calculating the mean.
QUESTION
What is the mean of the data set {10; 20; 30; 40; 50}?
SOLUTION
Step 1: Calculate the sum of the data

10+20+30+40+50=150

Step 2: Divide by the number of values in the data set to get the mean
Since there are 5 values in the data set, the mean is:

mean=150/5=30
More Related Questions and the Solutions

Q1. Calculate the mean of the following data set: {9; 14; 9; 14; 8; 8; 9; 8; 9; 9}. Round your answer to 1 decimal place.
solution:

The mean is: 9.7.

Q2. Find the mean of the natural numbers from 3 to 12.
solution:
Numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Here n=10

Q3. Find the mean of the following set of numbers:
(i) 6, 9, 11, 12 and 7
(ii) 11, 14, 23, 26, 10, 12, 18 and 6
solution:
(i)

Here n=5

(ii)

Here n=8

Q4. Marks obtained (in mathematics) by 9 student are given below:
60, 67, 52, 76, 50, 51, 74, 45 and 56

(a) find the arithmetic mean
(b) if marks of each student be increased by 4; what will be the new
value of arithmetic mean.
solution:
(a) Here n=9

(b)
If marks of each student be increased by 4 then new arithmetic mean will be=59+4=63

Q5. (a) Find the mean of 7, 11, 6, 5, and 6
(b) If each number given in (a) is diminished by 2, find the new value
of mean.
solution:
(a) The mean of 7, 11, 6, 5 and 6

(b)
If we subtract 2 from each number, then the mean will be 7-2=5

Q6. If the mean of 6, 4, 7, ‘a‘ and 10 is 8. Find the value of ‘a
solution:
No. of terms=5
Mean=8
Sum of numbers=8×5=40. (i)
But, sum of numbers=6+4+7+a+10=27+a … (ii)
From (i) and
(ii) 27+a=40→a=13

Q7. If 69.5 is the mean of 72, 70, ‘x‘, 62, 50, 71, 90, 64, 58 and 82, find the value of ‘x‘.
solution:
No. of terms =10
Mean =69.5
Sum of the numbers =69.5×10=695 … (i)
But sum of numbers =72+70+x+62+50+71+90+64+58+82=619+x … (ii)
from (i) and (ii) 619+x=695→x=76

Q8. In a Mathematics class, 23 learners completed a test out of 25 marks. Here is a list of their results:

14; 10; 23; 21; 11; 19; 13; 11; 20; 21; 9; 11; 17; 17; 18; 14; 19; 11; 24; 21; 9; 16; 6.

Calculate the mean of this data.
Solution:

Q9. Three data values are represented as follows: p+1;p+2;p+9. Find the mean in terms of p.
solution:

Miscellaneous Applications

Q10. A group of 10 friends each have some stones. They work out that the mean number of stones they have is 6. Then 7 friends leave with an unknown number (x) of stones. The remaining 3 friends work out that the mean number of stones they have left is 12.33.
When the 7 friends left, how many stones did they take with them?
solution:
If the mean number of stones the group originally had was 6 then the total number of stones must have been:

number of stones (before)=mean × group size

=(6)×(10)=60

We are then told that 7 friends leave and thereafter the mean number of stones left is 12.33. Now we can work out the remaining number of stones.

number of stones (after)=mean × group size

=(12.33)×(3)=37

Now we can calculate how many stones were taken by the 7 friends who left the group.
number of stones removed (x)=items before – items after

=(60)-(37)=23

Q11. A group of 9 friends each have some coins. They work out that the mean number of coins they have is 4. Then 5 friends leave with an unknown number (x) of coins. The remaining 4 friends work out that the mean number of coins they have left is 2.5.
When the 5 friends left, how many coins did they take with them?
solution:
If the mean number of coins the group originally had was 4 then the total number of coins must have been:

number of coins (before)=mean × group size

(4)×(9)=36

We are then told that 5 friends leave and thereafter the mean number of coins left is 2.5. Let us work out the remaining number of coins.

number of coins (after)=mean × group size

=(2.5)×(4)=10

Now we can calculate how many coins were taken by the 5 friends who left the group.
number of coins removed (x)= items before – items after

=(36)-(10)=26
Q12. A group of 9 friends each have some marbles. They work out that the mean number of marbles they have is 3. Then 3 friends leave with an unknown number (x) of marbles. The remaining 6 friends work out that the mean number of marbles they have left is 1.17.
When the 3 friends left, how many marbles did they take with them?
solution:
If the mean number of marbles the group originally had was 3 then the total number of marbles must have been:

number of marbles (before)=mean × group size

=(3)×(9)=27

We are then told that 3 friends leave and thereafter the mean number of marbles left is 1.17. Let us work out the remaining number of marbles.

number of marbles (after)=mean × group size

=(1.17)×(6)=7

Now we can calculate how many marbles were taken by the 3 friends who left the group.
number of marbles removed (x)= items before – items after

(27)-(7)=20
Q13. In the first of a series ofjars, there is 1 sweet. In the second jar, there are 3 sweets. The mean number of sweets in the first two jars is 2.
a) If the mean number of sweets in the first three jars is 3, how many sweets are there in the third jar?
solution:
Let n3 be the number of sweets in the third jar:

b) If the mean number of sweets in the first four jars is 4, how many sweets are there in the fourth jar? solution: Let n4 be the number of sweets in the fourth jar:

Q14. Find a set of five ages for which the mean age is 5, the modal age is 2 and the median age is 3 years.
solution:
Let the five different ages be x1, x2, x3, x4 and x5. Therefore the mean is:

The median value is at position 3, therefore x3=3.
The mode is the age that occurs most often. We have 5 ages to work with and we know one of the ages is 3 (from the median). So the ordered data set is: {x1;x2;3;x4;x5 } (remember that we always calculate mean, mode and median using the ordered data set). We are told that the mode is 2. Looking at the ordered data set we see that either x1 or x2 must be 2 (x4 and x5 cannot be 2 as that would make the data set unordered). However, if only one of these values is 2 then the mode will not be 2. Therefore x1=x2=2.
So we can now update our calculation of the mean:

2+2+3+x4+x5=25
18=x4+x5

x4 and x5 can be any numbers that add up to 18 and are not the same (ifthey were the same then the mode would not be 2), so 12 and 6 or 8 and 10 or 3 and 15, etc.
Possible data sets:

Data set 1: {2; 2; 3; 4; 14}
Data set 2: {2; 2; 3; 5; 13}
Data set 3: {2; 2; 3; 6; 12}
Data set 4: {2; 2; 3; 7; 11}
Data set 5: {2; 2; 3; 8; 10}

Note that the set of ages must be ordered, the median value must be 3 and there must be 2 ages of 2.

Q15. Four friends each have some marbles. They work out that the mean number of marbles they have is 10. One friend leaves with 4 marbles. How many marbles do the remaining friends have together?
solution:
Let the number of marbles per friend be x1, x2, x3 and x4.

One friend leaves:

x1+x2+x3=40-4
x1+x2+x3=36

Therefore the remaining friends have 36 marbles.

Q16. A group of 7 friends each have some sweets. They work out that the mean number of sweets they have is 6. Then 4 friends leave with an unknown number (x) of sweets. The remaining 3 friends work out that the mean number of sweets they have left is 10.67.
When the 4 friends left, how many sweets did they take with them?
solution:
If the mean number of sweets the group originally had was 6 then the total number of sweets must have been:

number of sweets (before)=mean × group size

=6×7=42

We are then told that 4 friends leave and thereafter the mean number of sweets left is 10.67. Let us work out the remaining number of sweets.

number of sweets (after)=mean × group size

=10.67×3=32

Now we can calculate how many sweets were taken by the 4 friends who left the group.
number of sweets removed (x)= items before – items after

=42-32=10
Q17. A group of 10 friends each have some sweets. They work out that the mean number of sweets they have is 3. Then 5 friends leave with an unknown number (x) of sweets. The remaining 5 friends work out that the mean number of sweets they have left is 3.
When the 5 friends left, how many sweets did they take with them?
solution:
If the mean number of sweets the group originally had was 3 then the total number of sweets must have been:

number of sweets (before) = mean × group size

=3×10=30
We are then told that 5 friends leave and thereafter the mean number of sweets left is 3. Let us work out the remaining number of sweets.

number of sweets (after) = mean × group size

3×5=15

Now we can calculate how many sweets were taken by the 5 friends who left the group.
number of sweets removed (x)= items before – items after

=30-15=15
Q18. Five data values are represented as follows: 3x; x+2; x-3; x+4; 2x-5, with a mean of 30. Solve for x.
solution:

Q19. An engineering company has designed two different types of engines for motorbikes. The two different motorbikes are tested for the time (in seconds) it takes for them to accelerate from 0 km∙h-1 to 60 km∙h-1.

a) Which measure of central tendency should be used for this information?
solution:
Mean and mode. The mean will give us the average acceleration time, while the mode will give us the time that is most often obtained.
If we used the median we would not get any useful information as all the median tells us is what the central value is. The mean and mode provide more information about the data set as a whole.
b) Calculate the measure of central tendency that you chose in the previous question, for each motorbike.
solution:
We first sort the data.
Bike 1: {0.68; 0.71 ; 0.80; 0.87; 0.92; 1.00; 1.06; 1.09; 1.49; 1.55}.
Bike 2: {0.9; 0.9; 0.9; 0.9; 1.0; 1.0; 1.0; 1.0; 1.1; 1.1}.
Next we can calculate the mean for each bike:
mean bike 1 =

mean bike 2 =

For bike 1 the mean is 1.02 s and there is no mode, because there is no value that occurs more than once.
For bike 2 the mean is 1.0 s and there are two modes, 1.0 and 0.9.

c) Which motorbike would you choose based on this information? Take note of the accuracy of the numbers from each set of tests.
solution:
It would be difficult to choose. Although bike 1 appears to do better than bike 2 from the mean, the data for bike 2 is less accurate than that for bike 1 (it only has 1 decimal place). If we were to calculate the mean for bike 1 using only 1 decimal place we would get 0.9 s. This would make bike 2 better. Also bike 2 produces more consistent numbers. So bike 2 would likely be a good choice, but more information or more accurate information should be obtained.