Let the mean and variance of the frequency di | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The total frequency is $N = 4 + 4 + \alpha + \beta = 8 + \alpha + \beta$. The mean is given by $\bar{x} = \frac{\sum f_i x_i}{N} = 6$. $\frac{4(2)

Vedclass · Accepted Answer

$\frac{17}{3}$

Vedclass · Answer

(C) The total frequency is $N = 4 + 4 + \alpha + \beta = 8 + \alpha + \beta$. The mean is given by $\bar{x} = \frac{\sum f_i x_i}{N} = 6$. $\frac{4(2) + 4(6) + \alpha(8) + \beta(9)}{8 + \alpha + \beta} = 6$ $8 + 24 + 8\alpha + 9\beta = 48 + 6\alpha + 6\beta$ $2\alpha + 3\beta = 16 \quad \dots (i)$ The variance is $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2 = 6.8$. $\frac{4(4) + 4(36) + \alpha(64) + \beta(81)}{8 + \alpha + \beta} - 36 = 6.8$ $\frac{16 + 144 + 64\alpha + 81\beta}{8 + \alpha + \beta} = 42.8$ $160 + 64\alpha + 81\beta = 42.8(8 + \alpha + \beta) = 342.4 + 42.8\alpha + 42.8\beta$ $21.2\alpha + 38.2\beta = 182.4$ Multiplying by $10$: $212\alpha + 382\beta = 1824 \Rightarrow 106\alpha + 191\beta = 912 \quad \dots (ii)$ Solving $(i)$ and $(ii)$: From $(i)$,$\alpha = \frac{16 - 3\beta}{2} = 8 - 1.5\beta$. Substituting in $(ii)$: $106(8 - 1.5\beta) + 191\beta = 912$ $848 - 159\beta + 191\beta = 912 \Rightarrow 32\beta = 64 \Rightarrow \beta = 2$. Then $\alpha = 8 - 1.5(2) = 5$. Total frequency $N = 8 + 5 + 2 = 15$. New sum of observations when $x_3$ changes from $8$ to $7$: Original sum $= 6 	imes 15 = 90$. New sum $= 90 - (8 	imes 5) + (7 	imes 5) = 90 - 40 + 35 = 85$. New mean $= \frac{85}{15} = \frac{17}{3}$.

$x$	$2$	$6$	$8$	$9$
$f$	$4$	$4$	$\alpha$	$\beta$

Let the mean and variance of the frequency distribution be $6$ and $6.8$ respectively.

$x$ $2$ $6$ $8$ $9$

$f$ $4$ $4$ $\alpha$ $\beta$

If $x_{3}$ is changed from $8$ to $7$,then the mean for the new data will be:

Similar Questions

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 mortality in a town during $4$ quarters of a year due to various causes is given in the bar graph below. Based on this data,calculate the percentage increase in mortality in the third quarter compared to the second quarter.

If the median and the range of four numbers $\{x, y, 2x + y, x - y\}$,where $0 < y < x < 2y$,are $10$ and $28$ respectively,then the mean of the numbers is

The mean age of $25$ teachers in a school is $40 \text{ years}$. $A$ teacher retires at the age of $60 \text{ years}$ and a new teacher is appointed in his place. If now the mean age of the teachers in this school is $39 \text{ years}$,then the age (in years) of the newly appointed teacher is

If the coefficient of variation and standard deviation of a distribution are $50\%$ and $20$ respectively,what is its mean?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let the mean and variance of the frequency distribution be $6$ and $6.8$ respectively. $x$ $2$ $6$ $8$ $9$ $f$ $4$ $4$ $\alpha$ $\beta$ If $x_{3}$ is changed from $8$ to $7$,then the mean for the new data will be: