The mean and $S.D.$ of the marks of $200$ candidates were found to be $40$ and $15$ respectively. Later, it was discovered that a score of $40$ was wrongly read as $50$. The correct mean and $S.D.$ respectively are...
The mean and $S.D.$ of the marks of $200$ candidates were found to be $40$ and $15$ respectively. Later, it was discovered that a score of $40$ was wrongly read as $50$. The correct mean and $S.D.$ respectively are...
- A
$14.98, 39.95$
- B
$39.95, 14.98$
- C
$39.95, 224.5$
- D
None of these
Similar Questions
The variance $\sigma^2$ of the data is $ . . . . . .$
$x_i$
$0$
$1$
$5$
$6$
$10$
$12$
$17$
$f_i$
$3$
$2$
$3$
$2$
$6$
$3$
$3$
The variance $\sigma^2$ of the data is $ . . . . . .$
| $x_i$ | $0$ | $1$ | $5$ | $6$ | $10$ | $12$ | $17$ |
| $f_i$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
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Let $x_1, x_2,........,x_n$ be $n$ observations such that $\sum {{x_i}^2 = 300} $ and $\sum {{x_i} = 60} $ on value of $n$ among the following is
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Let $x _1, x _2, \ldots \ldots x _{10}$ be ten observations such that $\sum_{i=1}^{10}\left(x_i-2\right)=30, \sum_{i=1}^{10}\left(x_i-\beta\right)^2=98, \beta>2$ and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2\left( x _1-1\right)+4 \beta, 2\left( x _2-1\right)+$ $4 \beta, \ldots . ., 2\left(x_{10}-1\right)+4 \beta$, then $\frac{\beta \mu}{\sigma^2}$ is equal to :
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Suppose a class has $7$ students. The average marks of these students in the mathematics examination is $62$, and their variance is $20$ . A student fails in the examination if $he/she$ gets less than $50$ marks, then in worst case, the number of students can fail is
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Mean and variance of a set of $6$ terms is $11$ and $24$ respectively and the mean and variance of another set of $3$ terms is $14$ and $36$ respectively. Then variance of all $9$ terms is equal to
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