Hello, and welcome to today's mathematics lesson. Today, we are going to explore Chapter 19: Mean and Median. By the end of this lesson, you will understand how to calculate the mean and median of ungrouped data, and you will discover some fascinating properties that make working with averages much simpler.
Let us begin with the mean. The mean is what most people call the average. It is one of the most common ways to find a representative value for a set of numbers.
Imagine you have several observations: x₁, x₂, x₃, and so on, up to xₙ. There are n observations in total. The mean, which we denote by x̄, is calculated by adding all the observations together and then dividing by the total number of observations.
So the formula is: x̄ = (x₁ + x₂ + x₃ + ... + xₙ)/n. We can also write this more compactly as x̄ = (1/n) · Σx, where Σx simply means the sum of all the x values.
Let us work through an example together. Suppose we want to find the mean height of six boys whose heights are 146 centimetres, 154 centimetres, 153 centimetres, 160 centimetres, 157 centimetres, and 160 centimetres.
First, we add all the heights: 146 plus 154 plus 153 plus 160 plus 157 plus 160 equals 930 centimetres. Then we divide by 6, since there are six boys. 930 divided by 6 gives us 155 centimetres. Therefore, the mean height is 155 centimetres.
Here is another example. Find the mean of all prime numbers between 20 and 50. The prime numbers in this range are 23, 29, 31, 37, 41, 43, and 47. There are 7 prime numbers. Their sum is 251. Dividing 251 by 7 gives us 35 6/7. So the mean is 35 6/7.
Now, let us explore some remarkable properties of the mean that can save you a lot of calculation time.
Property one: The sum of deviations of all observations from their mean is always zero. This means if you take each observation, subtract the mean from it, and add up all these differences, you will get zero. The positive deviations exactly balance the negative deviations.
For example, consider the first five even natural numbers: 2, 4, 6, 8, and 10. Their mean is 6. The deviations are: 2 minus 6 equals negative 4, 4 minus 6 equals negative 2, 6 minus 6 equals 0, 8 minus 6 equals 2, and 10 minus 6 equals 4. Adding these: negative 4, negative 2, plus 0, plus 2, plus 4 equals zero. This property always holds true.
Property two: If you add a constant to every observation, the mean increases by that same constant. So if each observation is increased by a, the new mean becomes x̄ + a.
Property three: Similarly, if you subtract a constant from every observation, the mean decreases by that same constant. The new mean becomes x̄ − a.
Property four: If you multiply every observation by a constant, the mean is also multiplied by that constant. The new mean becomes a · x̄.
Property five: And if you divide every observation by a constant, the mean is divided by that same constant. The new mean becomes x̄/a.
Let us see these properties in action. Suppose the mean of ten numbers is 13.5. If each number is increased by 3, the new mean becomes 16.5. If each is decreased by 2, the new mean becomes 11.5. If each is multiplied by 4, the new mean becomes 54. And if each is divided by 5, the new mean becomes 2.7. These properties let you find new means instantly without recalculating everything from scratch.
Sometimes we need to find missing values or correct errors in data using the mean.
Here is a useful technique. If you know the mean of several observations, you can always find their total sum by multiplying the mean by the number of observations. This is because Σx = n · x̄.
For instance, if the mean of 40 observations was found to be 160, but one value was copied wrongly, we can find the correct mean. First, the incorrect total sum is 40 times 160, which equals 6400. The value 165 was wrongly copied as 125, so we need to add back the difference: 6400 minus 125 plus 165 equals 6440. The correct mean is 6440 divided by 40, which equals 161.
Similarly, if the mean of 15 observations is 200, and excluding one observation gives a mean of 198 for the remaining 14 observations, we can find the excluded value. The total of 15 observations is 15 times 200, which equals 3000. The total of 14 observations is 14 times 198, which equals 2772. Therefore, the excluded observation is 3000 minus 2772, which equals 228.
Now let us turn to the median, another important measure of central tendency.
The median is the middle value of a data set when it is arranged in order. Unlike the mean, the median is not affected by extremely high or extremely low values, which makes it very useful in certain situations.
To find the median, follow these steps. First, arrange all the data in ascending or descending order. Second, count the total number of observations, n. Third, apply the appropriate formula based on whether n is odd or even.
If n is odd, the median is simply the value of the ((n+1)/2)th term. This gives you the exact middle position.
If n is even, there is no single middle term. Instead, the median is the average of the two middle terms: the (n/2)th term and the (n/2 + 1)th term. So the formula is: median equals half of the sum of these two middle values.
Let us work through examples.
First, an odd number of observations. Find the median of: 18, 30, 39, 36, 28, 27, 31, 40, 33, 25, and 37. Arranged in ascending order: 18, 25, 27, 28, 30, 31, 33, 36, 37, 39, 40. There are 11 observations, so n equals 11. Since 11 is odd, the median is the ((11+1)/2) term, which is the 6th term. Counting to the sixth term: 18, 25, 27, 28, 30, 31. The median is 31.
Now, an even number of observations. Find the median of: 34, 47, 41, 52, 53, 56, 35, 49, 55, and 42. Arranged in ascending order: 34, 35, 41, 42, 47, 49, 52, 53, 55, 56. There are 10 observations, so n equals 10. Since 10 is even, we need the average of the 5th and 6th terms. The 5th term is 47, and the 6th term is 49. The median is half of the sum of 47 and 49, which equals half of 96, which is 48.
Here is a practical example. The weights of 12 students in kilograms are: 40, 61, 54, 50, 59, 37, 51, 41, 48, 62, 46, and 34. Arranged in order: 34, 37, 40, 41, 46, 48, 50, 51, 54, 59, 61, 62. With 12 observations, we average the 6th and 7th terms. The 6th term is 48, and the 7th term is 50. The median weight is half of the sum of 48 and 50, which equals 49 kilograms.
If we replace 62 kilograms with 35 kilograms, the ordered data becomes: 34, 35, 37, 40, 41, 46, 48, 50, 51, 54, 59, 61. Now the 6th term is 46 and the 7th term is 48. The new median is half of the sum of 46 and 48, which equals 47 kilograms. Notice how changing an extreme value affected the mean but only slightly shifted the median.
Sometimes we need to find an unknown value given the median. For example, if the median of fourteen observations arranged in ascending order is 20, and the 7th term is x + 2 while the 8th term is x + 4, we can set up an equation. The median equals half of the sum of the 7th term and the 8th term. So 20 equals half of (x + 2) + (x + 4). This simplifies to 20 equals half of 2x + 6, which equals x + 3. Therefore, x equals 17.
Let us recap the key takeaways from today's lesson.
First, the mean or average of ungrouped data is the sum of all observations divided by the number of observations. We write this as x̄ = Σx/n or x̄ = (1/n) · Σx.
Second, the sum of deviations of the mean from all observations always equals zero.
Third, adding, subtracting, multiplying, or dividing every observation by a constant changes the mean by that same operation.
Fourth, to find the median, always arrange the data in order first.
If n is odd, the median is the middle term. If n is even, the median is the average of the two middle terms.
Fifth, the mean is affected by every value in the data set, while the median depends only on the middle position and is more resistant to extreme values.
And sixth, you can use the relationship Σx = n · x̄ to find totals, correct errors, or solve for missing values.
That brings us to the end of our lesson on Mean and Median. Remember, statistics is not just about formulas. It is about understanding data and making sense of the world around us. Keep practising with different data sets, and soon these calculations will become second nature. Thank you for listening, and I look forward to seeing you in the next lesson.