If the mean of a set of observations x_1, x_2 | vedclass

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(B) Given that the mean of $x_1, x_2, \dots, x_{10}$ is $20$,we have: $\frac{x_1 + x_2 + \dots + x_{10}}{10} = 20 \implies \sum_{i=1}^{10} x_i = 200$

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(B) Given that the mean of $x_1, x_2, \dots, x_{10}$ is $20$,we have: $\frac{x_1 + x_2 + \dots + x_{10}}{10} = 20 \implies \sum_{i=1}^{10} x_i = 200$ The new set of observations is $(x_1 + 4), (x_2 + 8), \dots, (x_{10} + 40)$. The mean of the new set is: $	ext{Mean} = \frac{(x_1 + 4) + (x_2 + 8) + \dots + (x_{10} + 40)}{10}$ $	ext{Mean} = \frac{(x_1 + x_2 + \dots + x_{10}) + (4 + 8 + \dots + 40)}{10}$ $	ext{Mean} = \frac{\sum x_i}{10} + \frac{4(1 + 2 + \dots + 10)}{10}$ Using the sum of first $n$ natural numbers formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we get $\sum_{k=1}^{10} k = \frac{10 	imes 11}{2} = 55$. $	ext{Mean} = 20 + \frac{4 	imes 55}{10} = 20 + \frac{220}{10} = 20 + 22 = 42$.

$x_i$	$3$	$6$	$9$	$12$
$f_i$	$1$	$2$	$3$	$4$

Marks	$2$	$3$	$5$	$7$
Frequency	$(x+1)^2$	$2x-5$	$x^2-3x$	$x$

If the mean of a set of observations $x_1, x_2, \dots, x_{10}$ is $20$,then find the mean of $x_1 + 4, x_2 + 8, x_3 + 12, \dots, x_{10} + 40$.

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The mean of a set of numbers is $\bar{x}$. If each number is multiplied by $\lambda$,then the mean of the new set is

If the values $1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n}$ have frequencies $1, 2, 3, \dots, n$ respectively,what is their mean?

Find the mean of the following frequency distribution:

$x_i$ $3$ $6$ $9$ $12$

$f_i$ $1$ $2$ $3$ $4$

The number of observations in a group is $40$. If the average of the first $10$ observations is $4.5$ and the average of the remaining $30$ observations is $3.5$,then the average of the whole group is:

If for some positive $x \in R$,the frequency distribution of the marks obtained by $20$ students in a certain test is as follows:
Marks $2$ $3$ $5$ $7$
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Then the mean of the marks is:

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