Mean, also called the arithmetic mean or average, is obtained by dividing the sum of all the numbers by the quantity of the numbers.
Find the mean of 3, 4 and 5
`(3 + 4 + 5)/6 = 12/6 = 2`
The basic formula of Mean is
`bar(X) = (sum_(i=1)^N X_i)/N `
Where,
`bar(X)` (read as "x" bar) - is the symbol for mean
∑ (the Greek letter sigma) - is the symbol for sum
`Χ_i` - is the sum of the numbers
N- is the quantity of numbers
We find the mean by applying the formula,
`bar(X) = (sumx)/N `
`bar(X)` = `(5 + 10 + 15 + 20 + 30 + 90) /6`
`bar(X)` = ` 170 /6`
`bar(X)` = `28.33`
So, the mean is `28.33`
Data can be represented as ungrouped or grouped.
It is data obtained in original / raw form.
Marks obtained by 22 students in Science exam are as follows:
80, 72, 80, 75, 60, 40, 60, 44, 45, 47, 62, 46, 42, 42, 41,50, 51, 72, 73, 84, 87, 85
When we organize the raw data and put it in groups (known as classes), it is called grouped data.
When we put the same data in a more condensed form, and mention the frequency of each group, then such a table is called grouped frequency distribution table.
| Math scores | Frequency (f) |
|---|---|
| 80-89 | 5 |
| 70-79 | 4 |
| 60-69 | 3 |
| 50-59 | 2 |
| 40-49 | 8 |
The formula for the calculation of mean of ungrouped data is given below:
` bar(X)`= `(sum_(i=1)^n x_i)/n `=` (x_1+x_2+...+x_n)/n`
| Math scores | Frequency (f) |
|---|---|
| 1 | 60 |
| 2 | 47 |
| 3 | 73 |
| 4 | 69 |
| 5 | 56 |
| 6 | 82 |
| 7 | 51 |
| 8 | 63 |
| 9 | 77 |
| 10 | 90 |
This states that 10 students got 66.8 marks on an average.
The mean is calculated as:
`bar(X)= (sumx)/n `
`sumx= 60+47+73+69+56+82+51+63+77+90`
`bar(X)= 668/10`
`bar(X) = 66.8`
No. of observations, n = 7
z`sum_(i=1)^7 x_i= 50 + 100 + 150 + 200 + 250 + 300 + 350 `
`sum_(i=1)^7 x_i= 1400`
`barX= (sum_(i=1)^7 x_i)/n `
`barX= 1400/7`
`barX= 200`
Data represented in the form of frequency distribution is called grouped data. The following table illustrates the result of an experiment in which a dice was rolled 20 times.Below values were recorded every time.
1,2,3,4,5,6
| Math scores | Frequency (f) |
|---|---|
| 1 | 4 |
| 2 | 6 |
| 3 | 2 |
| 4 | 2 |
| 5 | 5 |
| 6 | 1 |
| Total = 20 |
We find the mean using the below formula:
`bar(X)= (sum_(i=1)^n fx_i)/(sum_(i=1)^n f_i)`
| Math scores | Frequency (f) |
|---|---|
| 1 | 4 |
| 2 | 6 |
| 3 | 2 |
| 4 | 2 |
| 5 | 5 |
| 6 | 1 |
| Total = 20 |
The formula to determine the arithmetic mean of grouped data is :
`bar(X)= (sum_(i=1)^n fx_i)/(sum_(i=1)^n f_i)`
We find the values of f.x (by multiplying f with x) as shown below:
| Number on the Dice (X) | Frequency (f) | f.X |
|---|---|---|
| 1 | 4 | 4 |
| 2 | 6 | 12 |
| 3 | 2 | 6 |
| 4 | 2 | 8 |
| 5 | 5 | 25 |
| 6 | 1 | 6 |
| `sum_(i=1)^6 f_i = 20` | `sum_(i=1)^6 fx_i = 61` |
Therefore
`bar(X)= (sum_(i=1)^6 fx_i)/(sum_(i=1)^6 f_i)`
`bar(X) = 61/( 20 )`
`bar(X) = 3.05 ~~ 3`
We round off the number to 3.
| Classes (X) | Frequency (f) |
|---|---|
| 0-9 | 7 |
| 10-19 | 4 |
| 20-29 | 8 |
| 30-39 | 5 |
| 40-49 | 9 |
| 50-59 | 7 |
| 60-69 | 10 |
| Total= 50 |
We need to find the midpoint of each of the class using the below formula.
Midpoint = `text(Upper limit + Lower Limit)/2`,
Using this formula, we find the midpoint of each group as shown in the table below. These values are denoted by x in the table.
| Number on the dice (X) | Frequency (f) | f.X |
|---|---|---|
| 1 | 4 | 4 |
| 2 | 6 | 12 |
| 3 | 2 | 6 |
| 4 | 2 | 8 |
| 5 | 5 | 25 |
| 6 | 1 | 6 |
| `sum_(i=1)^6 f_i = 20` | `sum_(i=1)^6 fx_i = 61` |
Next, we multiple each midpoint value with its corresponding value of frequency. The "f.x" values are recorded in a new column as shown in the following table.
| Number on the classes | Frequency(f) | f.X |
|---|---|---|
| 0-9 | 7 | 4.5 |
| 10-19 | 4 | 14.5 |
| 20-29 | 8 | 24.5 |
| 30-39 | 5 | 34.5 |
| 40-49 | 9 | 44.5 |
| 50-59 | 7 | 54.5 |
| 60-69 | 10 | 64.5 |
| Total= 50 |
Therefore,
`bar(X)= (sum_(i=1)^7 fx_i)/(sum_(i=1)^7 f_i) `
`= 1885/(50) = 37.7`
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