
The standard deviation of 10,16,10,16,10,10,16,16 is?
A) 4
B) 6
C) 3
D) 0
Answer
568.2k+ views
Hint: The Standard Deviation is a measure of how spread out numbers are. To calculate the standard deviation of those numbers, we just need to do the following steps.
1. Find the value of Mean of the given numbers(the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result
3. Then work out the mean of those squared differences.
4. Take the square root of the resulted value and we are done.
By converting the above statements into a formula we get
The formula for finding standard deviation(σ) is
\[\sigma = \sqrt {\dfrac{1}{N}\sum\limits_{i = 1}^N {{{({x_i} - \mu )}^2}} } \]
where N= Total number of data
\[{x_i}\]= value of each data
\[\mu \]= mean of the given data
Complete step by step answer:
From the given information, the data given is 10, 16, 10, 16, 10, 10, 16, 16
So N = 8
We know that\[\mu \]=$\dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{N}$ as it is the mean of all the data given.
Let us find the value of \[\mu \]using the data provided in the question.
\[\mu \]=$\dfrac{{10 + 16 + 10 + 16 + 10 + 10 + 16 + 16}}{8} = \dfrac{{104}}{8} = 13$
Now let us keep all the values known in the formula of standard deviation
$\begin{align}
& \sigma =\sqrt{\dfrac{1}{N}\sum\limits_{i=1}^{N}{{{({{x}_{i}}-\mu )}^{2}}}} \\
& =\sqrt{\dfrac{1}{8}[{{(10-13)}^{2}}+{{(16-13)}^{2}}+{{(10-13)}^{2}}+{{(16-13)}^{2}}+{{(10-13)}^{2}}+{{(10-13)}^{2}}+{{(16-13)}^{2}}+{{(16-13)}^{2}}]} \\
& =\sqrt{\dfrac{1}{8}[{{(-3)}^{2}}+{{(3)}^{2}}+{{(-3)}^{2}}+{{(3)}^{2}}+{{(-3)}^{2}}+{{(-3)}^{2}}+{{(3)}^{2}}+{{(3)}^{2}}]} \\
& =\sqrt{\dfrac{1}{8}[8{{(3)}^{2}}]} \\
\end{align}$ [since square of -3 and 3 are same]
= 3
Therefore the standard deviation of the given numbers 10, 16, 10, 16, 10, 10, 16, 16 is 3.
The correct option is C.
Note:
While finding the value of square root sometimes it can be positive or negative, check that according to the options. Do not make calculation mistakes. Standard deviation is the square root of variance , so check the question properly whether they have asked standard deviation or variance.
1. Find the value of Mean of the given numbers(the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result
3. Then work out the mean of those squared differences.
4. Take the square root of the resulted value and we are done.
By converting the above statements into a formula we get
The formula for finding standard deviation(σ) is
\[\sigma = \sqrt {\dfrac{1}{N}\sum\limits_{i = 1}^N {{{({x_i} - \mu )}^2}} } \]
where N= Total number of data
\[{x_i}\]= value of each data
\[\mu \]= mean of the given data
Complete step by step answer:
From the given information, the data given is 10, 16, 10, 16, 10, 10, 16, 16
So N = 8
We know that\[\mu \]=$\dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{N}$ as it is the mean of all the data given.
Let us find the value of \[\mu \]using the data provided in the question.
\[\mu \]=$\dfrac{{10 + 16 + 10 + 16 + 10 + 10 + 16 + 16}}{8} = \dfrac{{104}}{8} = 13$
Now let us keep all the values known in the formula of standard deviation
$\begin{align}
& \sigma =\sqrt{\dfrac{1}{N}\sum\limits_{i=1}^{N}{{{({{x}_{i}}-\mu )}^{2}}}} \\
& =\sqrt{\dfrac{1}{8}[{{(10-13)}^{2}}+{{(16-13)}^{2}}+{{(10-13)}^{2}}+{{(16-13)}^{2}}+{{(10-13)}^{2}}+{{(10-13)}^{2}}+{{(16-13)}^{2}}+{{(16-13)}^{2}}]} \\
& =\sqrt{\dfrac{1}{8}[{{(-3)}^{2}}+{{(3)}^{2}}+{{(-3)}^{2}}+{{(3)}^{2}}+{{(-3)}^{2}}+{{(-3)}^{2}}+{{(3)}^{2}}+{{(3)}^{2}}]} \\
& =\sqrt{\dfrac{1}{8}[8{{(3)}^{2}}]} \\
\end{align}$ [since square of -3 and 3 are same]
= 3
Therefore the standard deviation of the given numbers 10, 16, 10, 16, 10, 10, 16, 16 is 3.
The correct option is C.
Note:
While finding the value of square root sometimes it can be positive or negative, check that according to the options. Do not make calculation mistakes. Standard deviation is the square root of variance , so check the question properly whether they have asked standard deviation or variance.
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