Statistics practice · Easy

The average weight of 25 boys was calculated to be $78.4 \mathrm{~kg}$. It was later discovered that one weight was misread as $69 \mathrm{~kg}$ instead of $96 \mathrm{~kg}$. The correct average is

The mean of $n$ observations is $\bar x$. If one observation ${x_{n + 1}}$ is added, then the mean remains same. The value of ${x_{n + 1}}$ is

Mode of data $3,2,5,2,3,5,6,6,5,3,5,2,5$ is

The mean of the n observations ${x_1},{x_2},{x_3}, \ldots ,\;{x_n}$ be ${\bar x}$. Then , the mean of n observations $2{x_1} + 3,\;2{x_2} + 3,\;2{x_3} + 3,\; \ldots ,\;2{x_n} + 3$ is

The two regression lines for a bivariate data are $x+y+50=0$ and $2 x+3 y+k=0$. If $\bar{x}=0$, then $\bar{y}$ is

If mean and mode of a frequency distribution have same value, then coefficient of skewness is

If a variable takes values $0,\;1,\;2,\; \ldots ,n$ with frequencies $1,\;{\;^n}{C_1},{\;^n}{C_2}, \ldots ,{\;^n}{C_n},$ then the AM is

Variance of the data $2,4,5,6,8,17$ is 23.33. Then, variance of $4,8,10,12,16,34$ will be

The points scored by a basketball team in a series of matches are as follows:

$15,3,8,10,22,5,27,11,12,19,18,21,13,14$
Its median is

Out of two lines of regression given by $x+2 y=4$ and $2 x-3 y-5=0$, the regression of $x$ on $y$ is

The mean deviation from the mean of the set of observations, -1, 0, 4 is

For the frequency distribution
Variate $(x):{x_1}\,{x_2}\,{x_3}....\,{x_{15}}$
Frequency $(f):\,\,{f_1}\,{f_2}\,{f_3}......\,{f_{15}}$
Where $0 < {x_1} < {x_2} < {x_3} < ..... < {x_{15}} = 10$ and $\sum\limits_{i = 1}^{15} {{f_i}} > 0,$ the standard deviation cannot be

Coefficient of skewness for the values

Median $= 18.8,\;{Q_1} = 14.6,\;{Q_3} = 25.2$ is

The arithmetic mean of the first $n$ natural numbers is

Given 5 sample measurement of a part of machine as $6.33,6.37,6.36,6.32$ and 6.37 . The sample variance is

The weighted mean of first $n$ natural number if their weight are the same as the number is-

A school has four sections of chemistry in class XII having
40, 35, 45 and 42 students. The mean marks obtained in
chemistry test are 50, 60, 55 and 45 respectively for the
four sections, the over all average of marks per students is

If ${{G_1},{G_2}}$ are the geometric means of two series of observations and $G$ is the GM of the ratios of the corresponding observations then G is equal to

If r is Karl Pearson’s coefficient of correlation between two sets of variates, then

The mean income of a group of persons is $₹ 400$. Another group of persons has mean income $₹ 480.$ If the mean income of all the persons in the two groups together is $₹ 430$, then ratio of the number of persons in the groups is

The mean of the values $0,\;1,\;2,\;3,\;...\;n$ with the corresponding weights $^n{C_0},{\;^n}{C_1}, \ldots ,{\;^n}{C_n}$ respectively is

For the data $2,2,3,3,3,4,4,5,6,8,8,8,8$, the mean, median and mode are

In a class of 100 students, the average amount of pocket money is Rs 35 per student. If the average is Rs 25 for girls and Rs 50 for boys, then the number of girls in the class is

Coefficient of variation of two distribution are $50 \%$ and $60 \%$ and their standard deviation are 10 and 15, respectively. Then, difference of their arithmetic means is

If $\;\bar x = \bar y = 0,\;\sum {x_i}{y_i} = 12,\;{\sigma _X} = 2,\;{\sigma _Y} = 3\;and\;n = 10,$ then the coefficient of correlation is

The line of regression of $y$ on $x$ and $x$ on $y$ are respectively $y=x+5$ and $16 x-9 y=94$. If the variance of $y$ is 16 , then the variance of $x$ is

If $Var\left( x \right) = 8.25,$ $Var\left( y \right) = 33.96$ and $\mathrm{cov}\left( {x,y} \right) = 10.2$ then the correlation coefficient is

If the mean of a set of observations ${x_1},{x_2},\; \ldots {x_{10}}$ is 20, then the mean of ${x_1} + 4,\;{x_2} + 8,\; \ldots {x_{10}} + 40$ is

Mean and standard deviation from the following observations of marks of 5 students of a tutorial group (marks out of 25)8, 12, 13, 15, 22 are respectively

Assertion: The mean deviation of the data 2, 9, 9, 3, 6, 9, 4
from the mean is 2.57.

Reason: For individual observation, ${\text{Mean}}\,\,{\text{deviation}}\,\,\left( {\bar X} \right) = \frac{{\sum {\left| {{x_i} - \bar x} \right|} }}{n}$