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.
${x_i}$ | ${f_i}$ | ${f_i} = \frac{{{x_i} - 64}}{1}$ | ${y_i}^2$ | ${f_i}{y_i}$ | ${f_i}{y_i}^2$ |
$60$ | $2$ | $-4$ | $16$ | $-8$ | $32$ |
$61$ | $1$ | $-3$ | $9$ | $-3$ | $9$ |
$62$ | $12$ | $-2$ | $4$ | $-24$ | $48$ |
$63$ | $29$ | $-1$ | $1$ | $-29$ | $29$ |
$64$ | $25$ | $0$ | $0$ | $0$ | $0$ |
$65$ | $12$ | $1$ | $1$ | $12$ | $12$ |
$66$ | $10$ | $2$ | $4$ | $20$ | $40$ |
$67$ | $4$ | $3$ | $9$ | $12$ | $36$ |
$68$ | $5$ | $4$ | $16$ | $20$ | $80$ |
$100$ | $220$ | $0$ | $286$ |
Mean, $\bar x = A\frac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h = 64 + \frac{0}{{100}} \times 1 = 64 + 0 = 64$
Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 - \left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}^2} } \right)} } \right]$
$=\frac{1}{100^{2}}[100 \times 286-0]$
$=2.86$
$\therefore$ Standard deviation $(\sigma)=\sqrt{2.86}=1.69$
If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, …., (x_{50} - 4)^2$ 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 mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$ , respectively by a student who took by mistake $50$ instead of $40$ for one observation. What are the correct mean and standard deviation?
Find the mean and variance for the first $n$ natural numbers
Let $a_1, a_2, \ldots . a_{10}$ be $10$ observations such that $\sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50$ and $\sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \cdot \mathrm{a}_{\mathrm{j}}=1100$. Then the standard deviation of $a_1, a_2, \ldots, a_{10}$ is equal to :