In the previous section, we completed a discussion on the methods for calculating standard deviation. In this section, we will see how standard deviation can be used for the analysis of data.
Let us see an example where we use standard deviation to compare two data sets. It can be written in 7 steps:
1. Two different factories produce cement. They are named as Cement A and Cement B.
• We want to compare the strengths of the two cements.
2. Making cubes:
♦ Using Cement A, we make 30 concrete cubes.
♦ Using Cement B also, we make 30 concrete cubes.
3. Testing the cubes:
• We place each cube in a compression testing machine and apply a gradually increasing load. We note down the load at which the cube just begins to fail. (some images can be seen here)
• Let the expected load be 25.
4. Differences in the loads:
• For cement A, most of the thirty loads are close to 25. The readings are like: 23, 25, 27, 24 etc.,
• For cement B also, most of the thirty loads are close to 25. But there are some extreme values like, 12, 17, 31 etc.,
5. Comparing the mean values.
♦ For cement A, the mean ($\bar{x}_a$) of it’s thirty values is 24.8.
♦ For cement B also, the mean ($\bar{x}_b$) of it’s thirty values is 24.8.
6. Based on the mean values, we are tempted to conclude that, both A and B are of the same quality.
7. But in step (4), we saw that B has some extreme values.
• So standard variation (σB) of B will be larger than standard variation (σA) of A.
• Greater σ means, greater dispersion. So cement B has greater dispersion from the expected value.
• We can conclude that, cement A is of better quality.
(Several different types of tests must be carried out for assessing the quality of materials like cement, steel, sand, electronic capacitors etc., Statistics help us to analyze test results. We will see them in higher classes)
• The above example demonstrates how two data sets can be compared.
• Now let us see another case. It can be written in 6 steps:
1. We are given two data sets.
• The first of those two sets contains heights of 45 students.
• The other set contains weights of those 45 students.
• We want to compare the two sets.
2. Calculating standard deviations:
♦ Let the standard deviation of the first set be σ1
♦ Let the standard deviation of the second set be σ2
3. Comparing the two σ values:
♦ σ1 will be in cm.
♦ σ2 will be in kg.
• So we cannot directly compare the two σ values.
4. In such cases, we use coefficient of variation.
• It is calculated using the formula:
$\text{C.V}~=~\frac{\sigma}{\bar{x}} \times 100$
• Where:
♦ C.V is the coefficient of variation of the data set.
♦ σ is the standard deviation of that data set.
♦ $\bar{x}$ is the mean of that data set.
5. We see that:
♦ σ comes in the numerator
♦ $\bar{x}$ comes in the denominator.
• Since the above σ and $\bar{x}$ are related to the same data set, both will have the same units. So the units will get canceled. We can say that, C.V has no units.
6. In our present example:
• Both (C.V)1 and (C.V)2 will not have any units. They are just numbers. So we can compare them.
• A data set having larger C.V is said to be more dispersed than the other.
• A data set having smaller C.V is said to be more consistent than the other.
In some cases, the two data sets will have the same units. But the mean may be different. In such cases also, calculation of C.V becomes necessary.
Solved example 15.14
The following table shows the number of workers of two factories and their wages.
 |
| Table 15.27 |
Find the factory in which, wages have greater variability?
Solution:
1. Consider factory A
• There are 5000 workers in this factory. So the data set will contain 5000 wages.
• The mean of those 5000 wages can be easily calculated.
♦ Mean is given as: Rs 2500
• Once the mean is calculated, variance can be easily calculated.
♦ Variance is given as: 81
♦ So standard deviation = √81 = Rs 9
2. Consider factory B
• There are 6000 workers in this factory. So the data set will contain 6000 wages.
• The mean of those 6000 wages can be easily calculated.
♦ Mean is given as: Rs 2500
• Once the mean is calculated, variance can be easily calculated.
♦ Variance is given as: 100
♦ So standard deviation = √100 = Rs 10
3. Both data sets have the same mean, which is Rs 2500.
4. Based on the above steps, we can write:
• For A:
Dispersion of wages from the mean of 2500 is 9.
• For B:
Dispersion of wages from the mean of 2500 is 10.
5. Thus we get:
Factory B has greater dispersion (variability) of wages than Factory A.
Solved example 15.15
The C.V of two distributions are 60 and 70. Their standard deviations are 21 and 16 respectively. What are their arithmetic means.
Solution:
1. We have the formula:
$\text{C.V}~=~\frac{\sigma}{\bar{x}} \times 100$
• Where:
♦ C.V is the coefficient of variation of the data set.
♦ σ is the standard deviation of that data set.
♦ $\bar{x}$ is the mean of that data set.
2. For the first set, we can write:
$\begin{array}{ll}
{}&{60}
& {~=~}& {\frac{21}{\bar{x}} \times 100}
&{} \\
{\Rightarrow}&{\bar{x}}
& {~=~}& {\frac{21}{60} \times 100}
&{} \\
{}&{}
& {~=~}& {\frac{7}{20} \times 100}
&{} \\
{}&{}
& {~=~}& {7 \times 5}
&{} \\
{}&{}
& {~=~}& {35}
&{} \\
\end{array}$
3. For the second set, we can write:
$\begin{array}{ll}
{}&{70}
& {~=~}& {\frac{16}{\bar{x}} \times 100}
&{} \\
{\Rightarrow}&{\bar{x}}
& {~=~}& {\frac{16}{70} \times 100}
&{} \\
{}&{}
& {~=~}& {\frac{8}{35} \times 100}
&{} \\
{}&{}
& {~=~}& {\frac{8}{7} \times 20}
&{} \\
{}&{}
& {~=~}& {22.85}
&{} \\
\end{array}$
Solved example 15.16
The following values are calculated in respect of heights and weights of the students of a section of class XI:
 |
| Table 15.28 |
Can we say that the weights show greater variation than the heights?
Solution:
1. Consider the data set related to heights.
• Let there be n students. So the data set will contain n heights.
• The mean of those n heights can be easily calculated.
♦ Mean is given as: 162.6 cm
• Once the mean is calculated, variance can be easily calculated.
♦ Variance is given as: 127.69 cm2.
♦ So standard deviation = √127.69 = 11.3 cm.
2. Consider the data set related to weights.
• There are n students. So the data set will contain n weights.
• The mean of those n weights can be easily calculated.
♦ Mean is given as: 52.36 kg
• Once the mean is calculated, variance can be easily calculated.
♦ Variance is given as: 23.1361 kg2.
♦ So standard deviation = √23.1361 = 4.81 kg.
3. Comparing the results in (1) and (2), we see that, standard deviation is greater for heights.
• But we cannot make such a comparison. This is because, one data set is for heights and the other is for weights.
• To get rid of the units, we must calculate C.V of each data set.
4. We have the formula:
$\text{C.V}~=~\frac{\sigma}{\bar{x}} \times 100$
• Where:
♦ C.V is the coefficient of variation of the data set.
♦ σ is the standard deviation of that data set.
♦ $\bar{x}$ is the mean of that data set.
5. For the heights, we can write:
$\begin{array}{ll}
{}&{\text{C.V}}
& {~=~}& {\frac{11.3}{162.6} \times 100}
&{} \\
{}&{}
& {~=~}& {0.06949 \times 100}
&{} \\
{}&{}
& {~=~}& {6.95}
&{} \\
\end{array}$
6. For the weights, we can write:
$\begin{array}{ll}
{}&{\text{C.V}}
& {~=~}& {\frac{4.81}{52.36} \times 100}
&{} \\
{}&{}
& {~=~}& {0.0918 \times 100}
&{} \\
{}&{}
& {~=~}& {9.18}
&{} \\
\end{array}$
7. We see that:
C.V is greater for weights.
• So we can write:
Weights show greater variability than heights.
Link to a few more solved examples is given below:
In the next section, we will see some miscellaneous examples.
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