Mean using working (Assumed) Mean
Assumed mean is a method of calculating the arithmetic men and standard deviation of a data set. It simplifies calculation.
Example
The masses to the nearest kilogram of 40 students in the form 3 class were measured and recorded in the table below. Calculate the mean mass
| Mass (kg) | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Number of Employees | 2 | 0 | 1 | 2 | 3 | 2 | 5 | 6 | 7 | 5 | 3 | 2 | 1 | 1 |
Solution
We are using assumed mean of 53
Here is the data you provided in a table format:
| Mass (x) kg | t = x – 53 | f (Frequency) | ft (f * t) |
|---|---|---|---|
| 47 | -6 | 0 | 0 |
| 48 | -5 | 1 | -5 |
| 49 | -4 | 2 | -8 |
| 50 | -3 | 3 | -9 |
| 51 | -2 | 2 | -4 |
|
52 |
-1 | 5 | -5 |
| 53 | 0 | 6 | 0 |
| 54 | 1 | 7 | 7 |
|
55
|
2 | 14 | 28 |
| 56 | 3 | 5 | 15 |
| 57 | 4 | 3 | 12 |
| 58 | 5 | 2 | 10 |
| 59 | 6 | 1 | 6 |
| 60 | 7 | 1 | 7 |
| Σf = 40 | Σft = 40 |
Mean of t = ∑ 𝑓 𝑡
∑𝑓 = 40
40 = 1
Mean of x = 53 + mean of t
= 53 + 1
= 54
Mean of grouped data
The masses to the nearest gram of 100 eggs were as follows
Here is the data you provided in a table format:
| Marks Range | Frequency |
|---|---|
| 100 – 103 | 1 |
| 104 – 107 | 15 |
| 108 – 111 | 42 |
| 112 – 115 | 31 |
| 116 – 119 | 8 |
| 120 – 123 | 3 |
Find the mean mass
Solution
Let use a working mean of 109.5.
| Class Range | Mid-point (x) | t = x – 109.5 | Frequency (f) | ft = f * t |
|---|---|---|---|---|
| 100 – 103 | 101.5 | -8 | 1 | -8 |
| 104 – 107 | 105.5 | -4 | 15 | -60 |
| 108 – 111 | 109.5 | 0 | 42 | 0 |
| 112 – 115 | 113.5 | 4 | 31 | 124 |
| 116 – 119 | 117.5 | 8 | 8 | 64 |
| 120 – 123 | 121.5 | 12 | 3 | 36 |
| Σf = 100 | Σft = 156 |
Mean of t = 156
100 = 1.56
Therefore,mean of x = 109.5 + mean of t
= 109.5 + 1.56
= 111.06 g
To get the mean of a grouped data easily, we divide each figure by the class width after substracting the assumed mean.Inorder to obtain the mean of the original data from the mean of the new set of data, we will have to reverse the steps in the following order;
- Multiply the mean by the class width and then add the working mean.
Example
The example above to be used to demonstrate the steps
| Class Range | Mid-point (x) | t = x – 109.5 | Frequency (f) | ft = f * t |
|---|---|---|---|---|
| 100 – 103 | 101.5 | -2 | 1 | -2 |
| 104 – 107 | 105.5 | -1 | 15 | -15 |
| 108 – 111 | 109.5 | 0 | 42 | 0 |
| 112 – 115 | 113.5 | 1 | 31 | 31 |
| 116 – 119 | 117.5 | 2 | 8 | 16 |
| 120 – 123 | 121.5 | 3 | 3 | 9 |
| Σf = 100 | Σft = 39 |
𝑡̅ = ∑𝑓𝑡
∑𝑓 = 39
100
= 0.39
Therefore 𝑥̅ = 0.39 x 4 + 109.5
= 1.56 + 109.5
= 111.06 g
Quartiles, Deciles and Percentiles
A median divides a set of data into two equal part with equal number of items.
Quartiles divides a set of data into four equal parts.The lower quartile is the median of the bottom half.The upper quartile is the median of the top half and the middle coincides with the median of the whole set of data.
Deciles divides a set of data into ten equal parts.Percentiles divides a set of data into hundred equal parts.
Note;
For percentiles deciles and quartiles the data is arranged in order of size.
Example
| Height (cm) | Frequency |
|---|---|
| 145 – 149 | 2 |
| 150 – 154 | 5 |
| 155 – 159 | 16 |
| 160 – 164 | 9 |
| 165 – 169 | 5 |
| 170 – 174 | 2 |
| 175 – 179 | 1 |
Calculate the ;
a.) Median height
b.) I.)Lower quartile
ii) Upper quartile
c.) 80th percentile
Solution
I. There are 40 students. Therefore, the median height is the average of the heights of the 20th and 21st students.
| Class Range | Frequency | Cumulative Frequency |
|---|---|---|
| 145 – 149 | 2 | 2 |
| 150 – 154 | 5 | 7 |
| 155 – 159 | 16 | 23 |
| 160 – 164 | 9 | 32 |
| 165 – 169 | 5 | 37 |
| 170 – 174 | 2 | 39 |
| 175 – 179 | 1 | 40 |
Both the 20th and 21st students falls in the 155 -159 class. This class is called the median class. Using the formula m = L + (𝑛
2−𝐶)𝑖
𝑓
(n divided by 2 minus C multiplied by i , divided by f)
Where L is the lower class limit of the median class
N is the total frequency
C is the cumulative frequency above the median class
I is the class interval
F is the frequency of the median class
Therefore;
Height of the 20th student = 154.5 + 13
16 𝑥 5
= 154.5 + 4.0625
=158.5625
Height of the 21st = 154.5 + 14
16 𝑥 5
= 154.5 + 4.375
=158.875
Therefore median height = 158.5625+158.875
2
= 158.7 cm
b.) (I ) lower quartile 𝑄1= L + (𝑛
4−𝐶)𝑖
𝑓
The 10th student fall in the in 155 – 159 class
𝑄1= 154.5 + (40
4 −7)5
16
= 154.5 + 0.9375
= 155.4375
(ii) Upper quartile 𝑄3= L + (3
4 𝑛 −𝐶)𝑖
𝑓
The 10th student fall in the in 155 – 159 class
𝑄3= 159.5 + (3
4 𝑥 40 −23)5
9
= 159.5 + 3.888
= 163.3889
Note;
The median corresponds to the middle quartile 𝑄2 or the 50th percentile
c.) 80
100 𝑥 40 = 32 the 32nd student falls in the 160 -164 class
𝑇ℎ𝑒 80𝑡ℎ 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒= L + ( 80
100 𝑛 −𝐶)𝑖
𝑓
= 159.5 + (32 −23)5
9
= 159.5 + 5
= 164.5
Example
Determine the upper quartile and the lower quartile for the following set of numbers
5, 10 ,6 ,5 ,8, 7 ,3 ,2 ,7 , 8 ,9
Solution
Arranging in ascending order
2, 3, 5,5,6, 7,7,8,8,9,10
The median is 7
The lower quartile is the median of the first half, which is 5.
The upper quartile is the median of the second half, which is 8.
Median from cumulative frequency curve
Graph for cumulative frequency is called an ogive. We plot a graph of cumulative frequency against the upper class limit.
Measure of Dispersion
Range
The difference between the highest value and the lowest value
Disadvantage
It depends only on the two extreme values
Interquartile range
The difference between the lower and upper quartiles. It includes the middle 50% of the values
Semi quartile range
The difference between the lower quartile and upper quartile divided by 2.It is also called the quartile
deviation.
Mean Absolute Deviation
If we find the difference of each number from the mean and find their mean , we get the mean Absolute deviation.
Variance
The mean of the square of the square of the deviations from the mean is called is called variance or mean deviation.
Example
| Deviation from Mean (d) | Frequency (fi) | fi * d |
|---|---|---|
| +1 | 1 | +1 |
| -1 | 1 | -1 |
| +6 | 36 | +216 |
| -4 | 16 | -64 |
| -2 | 4 | -8 |
| -11 | 121 | -1331 |
| +1 | 1 | +1 |
| +10 | 100 | +1000 |
Sum 𝑑2 = 1 + 1 + 36 + 16 + 4 + 121 + 1 + 100 = 280
Variance = ∑𝑑2
𝑁 = 280
8 = 35
The square root of the variance is called the standard deviation.It is also called root mean square deviation. For the above example its standard deviation =√35 = 5.9
Example
The following table shows the number of children per family in a housing estate
| Number of Children | Number of Families |
|---|---|
| 0 | 1 |
| 1 | 5 |
| 2 | 11 |
| 3 | 27 |
| 4 | 10 |
| 5 | 4 |
| 6 |
Calculate
a.) The mean number of children per family
b.) The standard deviation
Solution
| Number of Children (x) | Number of Families (f) | f⋅x | Deviation = x – m | f | |
|---|---|---|---|---|---|
| 0 | 1 | 0 | -3 | 9 | 9 |
| 1 | 5 | 5 | -2 | 4 | 20 |
| 2 | 11 | 22 | -1 | 1 | 11 |
| 3 | 27 | 81 | 0 | 0 | 0 |
| 4 | 10 | 40 | 1 | 1 | 10 |
| 5 | 4 | 20 | 2 | 4 | 16 |
| 6 | 2 | 12 | 3 | 9 | 18 |
| Σf = 60 | Σf= -40 |
a.) Mean = 180
60 = 3 𝑐ℎ𝑖𝑙𝑑𝑒𝑛
b.) Variance = ∑𝑓𝑑2
∑𝑓
= 84
60
= 1.4
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = √1.4 = 1.183
Example
The table below shows the distribution of marks of 40 candidates in a test
| Marks Range | Frequency (f) |
|---|---|
| 1 – 10 | 2 |
| 11 – 20 | 2 |
| 21 – 30 | 3 |
| 31 – 40 | 9 |
| 41 – 50 | 12 |
| 51 – 60 | 5 |
| 61 – 70 | 2 |
| 71 – 80 | 3 |
| 81 – 90 | 1 |
| 91 – 100 | 1 |
Calculate the mean and standard deviation.
| Marks | Midpoint (x) | Frequency (f) | fx | fx | d=x-m | f |
|---|---|---|---|---|---|---|
| 1 – 10 | 5.5 | 2 | 11.0 | -39.5 | 1560.25 | 3120.5 |
| 11 – 20 | 15.5 | 2 | 31.0 | -29.5 | 870.25 | 1740.5 |
| 21 – 30 | 25.5 | 3 | 76.5 | -19.5 | 380.25 | 1140.75 |
| 31 – 40 | 35.5 | 9 | 319.5 | -9.5 | 90.25 | 812.25 |
| 41 – 50 | 45.5 | 12 | 546.0 | 0.5 | 0.25 | 3.00 |
| 51 – 60 | 55.5 | 5 | 277.5 | 10.5 | 110.25 | 551.25 |
| 61 – 70 | 65.5 | 2 | 131.0 | 20.5 | 420.25 | 840.5 |
| 71 – 80 | 75.5 | 3 | 226.5 | 30.5 | 930.25 | 2790.75 |
| 81 – 90 | 85.5 | 1 | 85.5 | 40.5 | 1640.25 | 1640.25 |
| 91 – 100 | 95.5 | 1 | 95.5 | 50.5 | 2550.25 | 2550.25 |
Totals:
- Σf = 40
- Σf * x = 1800
- f = 15190
Mean 𝑥̅ = ∑𝑓𝑥
∑𝑓 = 1800
40 = 45 𝑚𝑎𝑟𝑘𝑠
Variance =∑𝑓𝑑2
∑𝑓 = 15190
40 = 379.75
= 379.8
Standard deviation = √379.8
= 19.49
Note;
Adding or subtracting a constant to or from each number in a set of data does not alter the value of the variance or standard deviation.
More formulas
The formula for getting the variance 𝑠2 𝑜𝑓 𝑎 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑥 𝑖𝑠
𝑑2 = ∑(𝑥 − 𝑥̅ )2𝑓
∑𝑓
= ∑𝑓𝑥2
∑𝑓 − (∑𝑓𝑥
∑𝑓 )2
Example
The table below shows the length in centimeter of 80 plants of a particular species of tomato
| Length Range | Frequency (f) |
|---|---|
| 152 – 156 | 12 |
| 157 – 161 | 14 |
| 162 – 166 | 24 |
| 167 – 171 | 15 |
| 172 – 176 | 8 |
| 177 – 181 | 7 |
Calculate the mean and the standard deviation
Solution
Let A = 169
| Length Range | Mid-point (x) | x – 169 | t = x – 169 | f | ft | ft² |
|---|---|---|---|---|---|---|
| 152 – 156 | 154 | -15 | -3 | 12 | -36 | 108 |
| 157 – 161 | 159 | -10 | -2 | 14 | -28 | 56 |
| 162 – 166 | 164 | -5 | -1 | 24 | -24 | 24 |
| 167 – 171 | 169 | 0 | 0 | 15 | 0 | 0 |
| 172 – 176 | 174 | 5 | 1 | 8 | 8 | 8 |
| 177 – 181 | 179 | 10 | 2 | 7 | 14 | 28 |
Summaries:
- Σf = 80
- Σft = -66
- Σft² = 224
𝑡̅ =−66
80 = −0.825
Therefore 𝑥̅ = −0.825 𝑥 5 + 169
= -4.125 + 169
= 164.875 ( to 4 s.f)
Variance of t =∑𝑓𝑡2
∑𝑓 − 𝑡2
= 224
80 − ( −0.825)2
= 2.8 – 0.6806
= 2.119
Therefore , variance of x = 2.119 x 52
= 52.975
= 52.98 ( 4 s.f)
Standard deviation of x = √52.98
= 7.279
= 7.28 (to 2 d.p)
End of topic
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