
Can a sample have a standard deviation of zero?
Answer
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Hint: In the above question, we have to determine if a sample or a set of some elements can have a standard deviation of zero or not. The standard deviation is not usually zero for most of the sets and samples but there are some exceptions where it can be found to be zero as well. The standard deviation for a sample or a set of \[n\] number of elements is denoted by \[\sigma \] and it is given by the formula written below as:
\[ \Rightarrow \sigma = \sqrt {\sum\limits_{i = 1}^n {\left( {{x_i} - \overline x } \right)} } \]
Where \[\overline x \] is the mean of the given sample or set, which is given by the formula:
\[ \Rightarrow \overline x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\]
Complete step-by-step answer:
We have to show if any sample or set of some elements can have its standard deviation equal to zero or not.
Let us assume a sample or set \[S\] which has \[n\] number of elements.
We can write the set \[S\] as,
\[ \Rightarrow S = \left\{ {{x_1},{x_2},{x_3}...{x_n}} \right\}\]
Now the mean of set \[S\] will be given by the ratio of sum of all elements of \[S\] to the total number of elements present in \[S\] .
Therefore, the mean \[\overline x \] is given by,
\[ \Rightarrow \overline x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\]
Now the standard deviation \[\sigma \] of the set \[S\] will be given by the formula,
\[ \Rightarrow \sigma = \sqrt {\sum\limits_{i = 1}^n {\left( {{x_i} - \overline x } \right)} } \]
Now since we want to check if the standard deviation \[\sigma \] can be equal to zero or not,
Hence we will put \[\sigma \] equal to zero in the above equation.
That gives us the new equation as,
\[ \Rightarrow 0 = \sqrt {\sum\limits_{i = 1}^n {\left( {{x_i} - \overline x } \right)} } \]
Squaring both sides, we get
\[ \Rightarrow 0 = \sum\limits_{i = 1}^n {\left( {{x_i} - \overline x } \right)} \]
This is only possible if,
\[ \Rightarrow 0 = \left( {{x_i} - \overline x } \right)\]
That gives us,
\[ \Rightarrow {x_i} = \overline x \]
Since \[{x_i}\] is an arbitrary element of the set \[S\] , that means every element of \[S\] is equal to the mean of \[S\] that is \[\overline x \] .
This situation is only possible if all elements are same i.e. if \[S\] is of the type,
\[ \Rightarrow S = \left\{ {a,a,a,a,....a} \right\}\]
Therefore, yes, the standard deviation of a sample can be zero if \[S = \left\{ {a,a,a,a,....a} \right\}\] .
Note: A standard deviation or \[\sigma \] is the measurement of how dispersed the data is in relation to the mean value of that sample. Low standard deviation means that the data are varying close around the mean, and high standard deviation indicates data are more spread out and are farther than each other.
For example, the means of the following two samples are the same: 14, 16, 15, 14, 16 and 1, 8, 14, 21, 31. However, the second sample is clearly more spread out than the first sample. If a set has a lower standard deviation, the values are not spread out too much as we can notice and compare the data in the two samples.
\[ \Rightarrow \sigma = \sqrt {\sum\limits_{i = 1}^n {\left( {{x_i} - \overline x } \right)} } \]
Where \[\overline x \] is the mean of the given sample or set, which is given by the formula:
\[ \Rightarrow \overline x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\]
Complete step-by-step answer:
We have to show if any sample or set of some elements can have its standard deviation equal to zero or not.
Let us assume a sample or set \[S\] which has \[n\] number of elements.
We can write the set \[S\] as,
\[ \Rightarrow S = \left\{ {{x_1},{x_2},{x_3}...{x_n}} \right\}\]
Now the mean of set \[S\] will be given by the ratio of sum of all elements of \[S\] to the total number of elements present in \[S\] .
Therefore, the mean \[\overline x \] is given by,
\[ \Rightarrow \overline x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\]
Now the standard deviation \[\sigma \] of the set \[S\] will be given by the formula,
\[ \Rightarrow \sigma = \sqrt {\sum\limits_{i = 1}^n {\left( {{x_i} - \overline x } \right)} } \]
Now since we want to check if the standard deviation \[\sigma \] can be equal to zero or not,
Hence we will put \[\sigma \] equal to zero in the above equation.
That gives us the new equation as,
\[ \Rightarrow 0 = \sqrt {\sum\limits_{i = 1}^n {\left( {{x_i} - \overline x } \right)} } \]
Squaring both sides, we get
\[ \Rightarrow 0 = \sum\limits_{i = 1}^n {\left( {{x_i} - \overline x } \right)} \]
This is only possible if,
\[ \Rightarrow 0 = \left( {{x_i} - \overline x } \right)\]
That gives us,
\[ \Rightarrow {x_i} = \overline x \]
Since \[{x_i}\] is an arbitrary element of the set \[S\] , that means every element of \[S\] is equal to the mean of \[S\] that is \[\overline x \] .
This situation is only possible if all elements are same i.e. if \[S\] is of the type,
\[ \Rightarrow S = \left\{ {a,a,a,a,....a} \right\}\]
Therefore, yes, the standard deviation of a sample can be zero if \[S = \left\{ {a,a,a,a,....a} \right\}\] .
Note: A standard deviation or \[\sigma \] is the measurement of how dispersed the data is in relation to the mean value of that sample. Low standard deviation means that the data are varying close around the mean, and high standard deviation indicates data are more spread out and are farther than each other.
For example, the means of the following two samples are the same: 14, 16, 15, 14, 16 and 1, 8, 14, 21, 31. However, the second sample is clearly more spread out than the first sample. If a set has a lower standard deviation, the values are not spread out too much as we can notice and compare the data in the two samples.
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