Calculate the standard deviation of the following data: 10, 20, 15, 8, 3, 4.
\[\begin{align}
& A.35.666 \\
& B.5.972 \\
& C.4.82 \\
& D.6.12 \\
\end{align}\]
Answer
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Hint: In this question, we need to find standard deviation of the given data. For this, we will first find the mean of the given data. For calculating mean, we will divide the sum of given terms by the number of terms. After that, we will take derivatives from mean in the frequency distribution table and then square those deviations. Using all the columns we will find standard deviation of the data using formula $\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}}$ where $\sum{{{x}^{2}}}$ is the sum of squares of deviation and N is the number of terms.
Complete step-by-step answer:
We are given data as: 10, 20, 15, 8, 3, 4.
Here, we are given six terms so number of terms = 6.
Let us find the mean of the given data using the formula, $\text{Mean}=\dfrac{\text{Sum of all terms}}{\text{Number of terms}}$.
So, let us find the sum of the given terms, we get:
$10+20+15+8+3+4=60$.
Therefore, $\text{Mean}=\dfrac{60}{6}=10$. Hence, $\overline{X}=10$.
Now, let us draw frequency distribution table having column X, x and ${{x}^{2}}$ where X is the list of terms, x is the derivative of X from mean $\left( \overline{X}=10 \right)$ and ${{x}^{2}}$ is the square of the deviation.
Frequency distribution table becomes:
Now, let us find the sum of the square deviations, we get:
$\text{Sum}=0+100+25+4+49+36=214$.
As we know, standard deviation is given by $\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}}$ where $\sigma $ is standard deviation, $\sum{{{x}^{2}}}$ is the sum of squared deviation from mean and N is the number of terms. Putting in the value of the given data, we get:
$\sigma =\sqrt{\dfrac{214}{6}}=\sqrt{35.66667}=5.972$.
Hence, required standard deviation is 5.972
Hence, option B is the correct answer.
So, the correct answer is “Option B”.
Note: Students should note the sum of deviation $\sum{x}$ is always zero when deviations are taken from mean. If we take deviations from some other term then formula of standard deviation is given by \[\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}-{{\left( \dfrac{\sum{x}}{N} \right)}^{2}}}\]. Since, we have taken deviation from mean and then $\sum{x}=0$. So, our formula is reduced to $\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}}$. Students should take care of the signs while calculating deviations.
Complete step-by-step answer:
We are given data as: 10, 20, 15, 8, 3, 4.
Here, we are given six terms so number of terms = 6.
Let us find the mean of the given data using the formula, $\text{Mean}=\dfrac{\text{Sum of all terms}}{\text{Number of terms}}$.
So, let us find the sum of the given terms, we get:
$10+20+15+8+3+4=60$.
Therefore, $\text{Mean}=\dfrac{60}{6}=10$. Hence, $\overline{X}=10$.
Now, let us draw frequency distribution table having column X, x and ${{x}^{2}}$ where X is the list of terms, x is the derivative of X from mean $\left( \overline{X}=10 \right)$ and ${{x}^{2}}$ is the square of the deviation.
Frequency distribution table becomes:
| X | $x=X-\overline{X}\left( \overline{X}=10 \right)$ | ${{x}^{2}}={{\left( X-\overline{X} \right)}^{2}}$ |
| 10 | 0 | 0 |
| 20 | 10 | 100 |
| 15 | 5 | 25 |
| 8 | -2 | 4 |
| 3 | -7 | 49 |
| 4 | -6 | 36 |
Now, let us find the sum of the square deviations, we get:
$\text{Sum}=0+100+25+4+49+36=214$.
As we know, standard deviation is given by $\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}}$ where $\sigma $ is standard deviation, $\sum{{{x}^{2}}}$ is the sum of squared deviation from mean and N is the number of terms. Putting in the value of the given data, we get:
$\sigma =\sqrt{\dfrac{214}{6}}=\sqrt{35.66667}=5.972$.
Hence, required standard deviation is 5.972
Hence, option B is the correct answer.
So, the correct answer is “Option B”.
Note: Students should note the sum of deviation $\sum{x}$ is always zero when deviations are taken from mean. If we take deviations from some other term then formula of standard deviation is given by \[\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}-{{\left( \dfrac{\sum{x}}{N} \right)}^{2}}}\]. Since, we have taken deviation from mean and then $\sum{x}=0$. So, our formula is reduced to $\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}}$. Students should take care of the signs while calculating deviations.
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