Statistics · Mathematics · TS EAMCET

The mean deviation about median of the numbers $3 x, 6 x, 9 x, \ldots .81 x$ is 91 , then $|x|=$

$$ \text { The coefficient of variation for the following data is } $$

$$ \begin{array}{llllll} \hline \text { Class interval } & 0-2 & 2-4 & 4-6 & 6-8 & 8-10 \\ \hline \text { Frequency } & 2 & 3 & 5 & 3 & 2 \\ \hline \end{array} $$

The mean deviation from the mean of the discrete data $2,3,5,7,11,13,17,19,22$ is

The probability distribution of a random variable $X$ is given below. Then, the standard deviation of $X$ is

$$ \begin{array}{llllll} \hline \boldsymbol{X}=\boldsymbol{x}_1 & 2 & 3 & 5 & 7 & 12 \\ \hline \boldsymbol{P}\left(\boldsymbol{X}=\boldsymbol{x}_1\right) & 3 k & k & k & 2 k & k \\ \hline \end{array} $$

The variance of the discrete data $3,4,5,6,7,8,10,13$ is

If a possion variate $X$ satisfies the relation $P(X=3)=P(X=5)$, then $P(X=4)=$

If the variance of the numbers $9,15,21, \ldots,(6 n+3)$ is $P$, then the variance of the first $n$ even numbers is

The mean deviation from the median for the following data is

$$ \begin{array}{cllllll} x_i & 2 & 9 & 8 & 3 & 5 & 7 \\ \hline f_i & 5 & 3 & 1 & 6 & 6 & 1 \\ \hline \end{array} $$

If three dice are thrown, then the mean of the sum of the numbers appearing on them is

The variance of the following continuous frequency distribution is

The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively. Later on it was found that one of the observations was taken as 50 in the place of 40 . If the wrong entry is replaced by the correct one, then the sum of the squares of all the observations is

The variance of 50 observations is 7 . Suppose that each observation in this data is multiplied by 6 and then 5 is subtracted from it. Then, the variance of that new data is

If $M$ and $\sigma^2$ represent respectively the mean deviation from the mean and the variance for the data $1,3,5,7$, $11,13,17,19,23$, then $3\left(\sigma^2-M\right)=$

If $X$ is a Poisson variate satisfying the condition $3 P(X=2)=P(X=4)$, then $P(X=6)=$

If the variance of the data $2,3,5,8,12$ is $\sigma^2$ and the mean deviation from the median for this data is $M$, then $\sigma^2-M=$

Assertion (A) The variance of the first $n$ odd natural numbers is $\frac{n^2-1}{3}$.

Reason (R) The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n\left(4 n^2-1\right)}{3}$.

Which of the following alternatives is correct?

Statement I The range of the ungrouped data does not change even if certain intermediate observations are removed

Statement II The value of the mean deviation of an ungrouped data about the median is always less than or equal to the value of the mean deviation computed about any other measure of central tendency

Statement III For a grouped data, range is approximated as the difference between the lower limit of the largest class and the upper limit of the smallest class

If 10 is the mean deviation of ' $n$ ' observations $x_1, x_2, x_3, \ldots, x_n$, then the mean deviation of the observations $\frac{2 x_1+5}{3}, \frac{2 x_2+5}{3}, \frac{2 x_3+5}{3}, \ldots . \frac{2 x_n+5}{3}$ is

There are $n$ observations and all of them are negative numbers. The ascending order of these observations is $x_1, x_2, \ldots . x_n$. If the signs of the first term and last term in that order are changed, then the range of the data is

The mean deviation from the mean for the observations $1,3,5,7,11,13,17,19,23$ is

The mean deviation from the mean of the discrete data $1,3,4,7,11,18,29,47,78$ is

If $\bar{x}$ is the mean of $n$ observations $x_1, x_2, \ldots ., x_n$ then the mean of the absolute deviations of these observations from $\bar{x}$ is

In a discrete data $\frac{1 \text { th }}{4}$ of the observations are equal to $a$, another $\frac{1 \text { th }}{4}$ of the observations are equal to $-a$. Out of the remaining, half of them are equal to $b$ and the rest are equal to $-b$. If the variance of all the observations is $(a b)$, then

For the following distribution, the mean deviation about the median is

$$ \begin{array}{cccccccc} \hline x_i & 6 & 12 & 18 & 24 & 30 & 36 & 42 \\ \hline f_i & 4 & 7 & 9 & 18 & 15 & 10 & 5 \\ \hline \end{array} $$

The mean and standard deviation of 100 observations $x_1, x_2, \ldots, x_{100}$ were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. Then the correct value of $\sum_{i=1}^{100} x_i^2=$

The coefficient of variation of the first 5 prime numbers is

If $S_1$ and $S_2$ are the variances of the first $2 k$ and $k(k>1)$ natural numbers respectively, then ( $S_1 / S_2$ ) lies in the interval

The standard deviations of two sets of observations $X=\left\{x_i\right\}$ and $Y=\left\{y_i\right\}(i=1,2, \ldots, 100)$ are respectively 5 and 6 . If $\bar{x}, \bar{y}$ are their means and $\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600$, then the standard deviation of $Z=\left\{z_i / z_i=x_i-y_i\right)$ is

$$ \text { The variance of the following frequency distribution is } $$

$$ \begin{array}{ccccccc} \hline \text { Classes } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\ \hline \text { Frequency } & 11 & 29 & 18 & 4 & 5 & 3 \\ \hline \end{array} $$

The mean deviation about the mean of the following data is nearly

$$ \begin{array}{ccccccccc} \hline \text { Size }(x) & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\ \hline \text { Frequency }(f) & 3 & 3 & 4 & 14 & 7 & 4 & 3 & 4 \\ \hline \end{array} $$

Assertion (A) Variance of $4 x_1, 4 x_2, \ldots, 4 x_n$ is 16 times the variance of $x_1, x_2, x_3, \ldots, x_n$

Reason (R) If $y=a x+b$, then variance of $y$ is a $($ variance of $x)+b$

The correct option among the following is

If $\alpha, \beta$ are respectively the mean deviation about the mean and variance of the first five prime numbers, then the ordered pair ( $\alpha, \beta$ )

For the following frequency distribution, the variance is approximately equal to

$$ \begin{array}{cccccc} \hline \begin{array}{c} \text { Class } \\ \text { Interval } \end{array} & 0-5 & 5-10 & 10-15 & 15-20 & 20-25 \\ \hline \text { Frequency } & 4 & 1 & 10 & 3 & 2 \\ \hline \end{array} $$

If the mean of the discrete distribution $8,9,6,5, x, 4$, 6, 5 is 6 , then its standard deviation (nearest to two decimal places) is