Arithmetic Mean Exercises

a) The mean of a data set is given by the sum of the data divided by the total number of data points. Thus:

b)

We observe that if all the data is multiplied by , then the arithmetic mean is also multiplied by . Therefore, it is possible to represent this observation in the following property satisfied by the arithmetic mean:

with a constant.

We know from the start that:

Now, let's calculate the mean of the set of seven numbers and develop as follows:

Thus, the mean remains the same.

The table shows the variable and the number of times it repeats in the data set . For this reason, we must complete the table with the product of the variable times its absolute frequency in order to have the sum of all values that repeast times, and thus be able to finally add all of them and divide by the amount of data generated. Observe the formula:

Here is the numerical development:

61	5	305
64	18	1152
67	42	2814
70	27	1890
73	8	584
	100	6745

Then we just need to perform the division:

Reaching the desired result.

First of all, notice that now the data is not represented in the same way as before; we have intervals of values. In this case, what we do is calculate something called the class mark . This consists of finding the mean between the two values that defines the interval, for example:

and so on with the other intervals.

Once the calculation is done, we complete the table with the product of the variable times its absolute frequency to calculate the mean:

Interval
[10, 15)	12.5	3	37.5
[15, 20)	17.5	5	87.5
[20, 25)	22.5	7	157.5
[25, 30)	27.5	4	110
[30, 35)	32.5	2	65
		21	457.5

We calculate the sum of the variable times its absolute frequency which is and divide it by the total number of data points which is . Thus:

From the given data we can construct the following table:

1
2	32	64
3	35	105
4	33	132
5
6	35	210

The sum of the data in column gives us:

We set this equal to 200, which is one of the problem's conditions. Thus:

Simplifying we obtain our first equation:

Now we calculate the mean of the distribution and set it equal to 3.6, which is the value indicated in the problem:

Again simplifying we obtain our second equation:

Therefore we have the following system of equations:

We solve it by the elimination method:

Therefore:

Thus we conclude that:

and

We begin by calculating the :

Interval
[0, 5)	2.5	3
[5, 10)	7.5	5
[10, 15)	12.5	7
[15, 20)	17.5	8
[20, 25)	22.5	2
[25, ∞)	---	6
		31

And here we observe that the mean CANNOT be calculated because we cannot find the class mark of the last interval.

a) The mean of a data set is given by the sum of the data divided by the total number of data points. Thus:

b)

We observe that if we add to all the data, then the arithmetic mean also increases by . Therefore, it is possible to represent this observation in the following property satisfied by the arithmetic mean:

with a constant.

a)

Team 1:

Team 2:

b)

To calculate the mean for the game, , we calculate the mean of the means and . Thus:

For the mean of the data set to remain the same, the two added data points must satisfy that their mean equals . We know that one of them is . Let's call the second data point "". Therefore, what we seek is:

We solve this first-degree equation:

Thus, the second data point must be . As we can easily verify, the mean remains :

The first multiples of are:

Thus, their mean is:

To generalize the previous result, note that the n-th multiple of is . Thus, the mean is:

Taking the value in the expression above, we obtain the result from the first part of the exercise.