If the mean of the data:Class5 - 1010 - 1515 | vedclass

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(C) The mean of the data is given by $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.Mid-values $(x_i)$: $7.5, 12.5, 17.5, 22.5, 27.5, 32.5$.Total frequency

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$2x^2 - 19x - 10 = 0$

Vedclass · Answer

(C) The mean of the data is given by $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.Mid-values $(x_i)$: $7.5, 12.5, 17.5, 22.5, 27.5, 32.5$.Total frequency $\sum f_i = 2 + k + 28 + 54 + (k + 1) + 5 = 90 + 2k$.Sum of $f_i x_i = (2 	imes 7.5) + (k 	imes 12.5) + (28 	imes 17.5) + (54 	imes 22.5) + ((k + 1) 	imes 27.5) + (5 	imes 32.5) = 15 + 12.5k + 490 + 1215 + 27.5k + 27.5 + 162.5 = 1910 + 40k$.Given $\bar{x} = 21$,so $\frac{1910 + 40k}{90 + 2k} = 21$.$1910 + 40k = 21(90 + 2k) \Rightarrow 1910 + 40k = 1890 + 42k$.$2k = 20 \Rightarrow k = 10$.Testing $k = 10$ in the given equations:$A) 2(10)^2 - 23(10) - 10 = 200 - 230 - 10 = -40 
eq 0$.$B) 4(10)^2 - 35(10) + 24 = 400 - 350 + 24 = 74 
eq 0$.$C) 2(10)^2 - 19(10) - 10 = 200 - 190 - 10 = 0$.$D) 2(10)^2 - 35(10) + 98 = 200 - 350 + 98 = -52 
eq 0$.Therefore,$k = 10$ is a root of the equation $2x^2 - 19x - 10 = 0$.

Class	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$
Frequency	$2$	$k$	$28$	$54$	$k + 1$	$5$

$x_i$	$2$	$4$	$6$	$8$	$10$	$12$	$14$	$16$
$f_i$	$4$	$4$	$\alpha$	$15$	$8$	$\beta$	$4$	$5$

Class	$4-8$	$8-12$	$12-16$	$16-20$
Frequency	$3$	$\lambda$	$4$	$7$

Metric	Mill-$A$	Mill-$B$
No. of workers	$500$	$600$
Average daily wage (in rupees)	$186$	$175$
Variance of distribution of wages	$81$	$100$

If the mean of the data:
Class $5 - 10$ $10 - 15$ $15 - 20$ $20 - 25$ $25 - 30$ $30 - 35$
Frequency $2$ $k$ $28$ $54$ $k + 1$ $5$

is $21$,then $k$ is one of the roots of the equation:

Similar Questions

If the mean and standard deviation ($S$.$D$.) of the data $3, 5, 7, a, b$ are $5$ and $2$ respectively,then $a$ and $b$ are the roots of the equation:

If the mean and variance of the frequency distribution are $9$ and $15.08$ respectively,then the value of $\alpha^2+\beta^2-\alpha \beta$ is $............$.

$x_i$ $2$ $4$ $6$ $8$ $10$ $12$ $14$ $16$

$f_i$ $4$ $4$ $\alpha$ $15$ $8$ $\beta$ $4$ $5$

If the mean and the variance of the data are $\mu$ and $19$ respectively,then the value of $\lambda+\mu$ is:
Class $4-8$ $8-12$ $12-16$ $16-20$
Frequency $3$ $\lambda$ $4$ $7$

An analysis of monthly wages paid to the workers of two jute mills $A$ and $B$ gives the following data:

Metric Mill-$A$ Mill-$B$

No. of workers $500$ $600$

Average daily wage (in rupees) $186$ $175$

Variance of distribution of wages $81$ $100$

Then:

Two sets,each of $20$ observations,have the same standard deviation $5$. The first set has a mean $17$ and the second a mean $22$. Determine the standard deviation of the set obtained by combining the given two sets.

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If the mean of the data:Class$5 - 10$$10 - 15$$15 - 20$$20 - 25$$25 - 30$$30 - 35$Frequency$2$$k$$28$$54$$k + 1$$5$is $21$,then $k$ is one of the roots of the equation: