# Computations Involving the Mean, Sample Size, and Sum of a Data Set

In this lesson, we solve problems involving, the sample size, sum of a data set and it’s mean. Any two of these three quantities are given and we find the third unknown quantity using the relation between these 3 quantities.

**Formula**

$Mean = \frac{Sum \:of \:the\: data}{Number \:of \:data}$

Sum of the data = Mean × Number of data

$Number\: of\: data = \frac{Sum \:of \:the \:data}{Mean}$

The average of x and 3 is equal to the average of x, 6 and 9. Find x

1, 1, 4, 4, 5, 6, 7, 7, 10, 10

### Solution

**Step 1:**

Average of x and 3 = $\frac{(x+3)}{2}$

Average of x, 6, and 9 = $\frac{(x+6+9)}{3}$

**Step 2:**

Given $\frac{(x+3)}{2} = \frac{(x+15)}{3}$

Solving we get 3x + 9 = 2x + 30 or

3x – 2x = x = 30 – 9 = 21

**Step 3:**

So x = 21

7 consecutive even integers have an average of 48. Find the average of the greatest two of those integers.

### Solution

**Step 1:**

Let the consecutive even integers be

x – 6, x – 4, x – 2, x, x + 2, x + 4, x + 6

Their average = $\frac{(x – 6 + x – 4 + x – 2 + x + x + 2 + x + 4 + x + 6)}{7} = \frac{7x}{7}$ = 48. So X=48

**Step 2:**

So the numbers are 42, 44, 46, 48, 50, 52, 54

The average of the two greatest of these integers 52 and 54 is (52 + 54)/2 = 53