The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & 2 & 3 & 4 & 5 & 6 & 7 \\ f & 4 & 9 & 16 & 14 & 11 & 6 \\ \hline \end{array}$
Find the standard deviation.
The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & 2 & 3 & 4 & 5 & 6 & 7 \\ f & 4 & 9 & 16 & 14 & 11 & 6 \\ \hline \end{array}$
Find the standard deviation.
$\begin{array}{|c|c|c|c|c|} \hline x_{i} & f_{i} & d_{i}=x_{i}-4 & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 2 & 4 & -2 & -8 & 16 \\ \hline 3 & 9 & -1 & -9 & 9 \\ \hline 4 & 16 & 0 & 0 & 0 \\ \hline 5 & 14 & 1 & 14 & 14 \\ \hline 6 & 11 & 2 & 22 & 44 \\ \hline 7 & 6 & 3 & 18 & 54 \\ \hline \text { Total } & n=60 & & \Sigma f_{i}=37 & \Sigma f_{i} d_{i}^{2}=137 \\ \hline \end{array}$
$ \therefore \quad SD =\sqrt{\frac{\Sigma f_{i} d_{i}^{2}}{n}-\left(\frac{\Sigma f_{i} d_{i}}{n}\right)^{2}}=\sqrt{\frac{137}{60}-\left(\frac{37}{60}\right)^{2}}=\sqrt{2.2833-(0.616)^{2}} $
$=\sqrt{2.2833-0.3794}=\sqrt{1.9037}=1.38 $
Similar Questions
If the standard deviation of $0, 1, 2, 3, …..,9$ is $K$, then the standard deviation of $10, 11, 12, 13 …..19$ is
If the standard deviation of $0, 1, 2, 3, …..,9$ is $K$, then the standard deviation of $10, 11, 12, 13 …..19$ is
The $S.D$ of $15$ items is $6$ and if each item is decreased or increased by $1$, then standard deviation will be
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The data is obtained in tabular form as follows.
${x_i}$
$60$
$61$
$62$
$63$
$64$
$65$
$66$
$67$
$68$
${f_i}$
$2$
$1$
$12$
$29$
$25$
$12$
$10$
$4$
$5$
The data is obtained in tabular form as follows.
| ${x_i}$ | $60$ | $61$ | $62$ | $63$ | $64$ | $65$ | $66$ | $67$ | $68$ |
| ${f_i}$ | $2$ | $1$ | $12$ | $29$ | $25$ | $12$ | $10$ | $4$ | $5$ |
Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance
Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .
Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.
Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance
Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .
Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.
- [AIEEE 2012]