If the mean of x 1 and x 2 is M 1 and that of x 1- x 2- x 3- x 4 is M 2- then the mean of ax 1- ax 2- x 3- a - x 4- a isA- M 1- M 2-2B- aM 1- M 2-2-2C- 1-2 a - a 2-1 M 1- M 2-D- 1-2 a -2 a 2-1 M 1- M 2-

The correct option is C 1/2a[(a^2 1)M_1+M_2] Given that x_1+x_2/2=M_1⇒ x_1+x_2=2M_1 and x_1+x_2+x_3+x_4=4M_2 ⇒ x_3+x_4=4M_2 2M_1 Mean of ax_1,ax_2, x_3/a and ...

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Byju's Answer

Standard IX

Mathematics

Frequency Table & Its Average

If the mean o...

Question

If the mean of $x_{1}$ and $x_{2}$ is $M_{1}$ and that of $x_{1}, x_{2}, x_{3}, x_{4}$ is $M_{2}$ , then the mean of $a x_{1}, a x_{2}, \frac{x_{3}}{a}, \frac{x_{4}}{a}$ is

$\frac{M_{1} + M_{2}}{2}$

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$\frac{a M_{1} + (\frac{M_{2}}{2})}{2}$

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$\frac{1}{2 a} [(a^{2} - 1) M_{1} + M_{2}]$

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$\frac{1}{2 a} [2 (a^{2} - 1) M_{1} + M_{2}]$

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Solution

The correct option is C
$\frac{1}{2 a} [(a^{2} - 1) M_{1} + M_{2}]$

Given that $\frac{x_{1} + x_{2}}{2} = M_{1} \Rightarrow x_{1} + x_{2} = 2 M_{1}$
and $x_{1} + x_{2} + x_{3} + x_{4} = 4 M_{2}$
$\Rightarrow x_{3} + x_{4} = 4 M_{2} - 2 M_{1}$
Mean of $a x_{1}, a x_{2}, \frac{x_{3}}{a} a n d \frac{x_{4}}{a} i s$
$= \frac{a x_{1} + a x_{2} + \frac{x_{3}}{a} + \frac{x_{4}}{a}}{4} = \frac{a}{4} (x_{1} + x_{2}) + \frac{1}{4 a} (x_{3} + x_{4}) = \frac{a}{4} .2 M_{1} + \frac{1}{4 a} (4 M_{2} - 2 M_{1}) = \frac{a M_{1}}{2} + \frac{M_{2}}{a} - \frac{M_{1}}{2 a} = \frac{1}{2 a} [(a^{2} - 1) M_{1} + 2 M_{2}]$ .

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If the mean of x1 and x2 is M1 and that of x1, x2, x3, x4 is M2, then the mean of ax1, ax2, x3a, x4a is

The correct option is C 12a[(a2−1)M1+M2] Given that x1+x22=M1⇒x1+x2=2M1 and x1+x2+x3+x4=4M2 ⇒x3+x4=4M2−2M1 Mean of ax1,ax2,x3a and x4a is =ax1+ax2+x3a+x4a4=a4(x1+x2)+14a(x3+x4)=a4.2M1+14a(4M2−2M1)=aM12+M2a−M12a=12a[(a2−1)M1+2M2].

If the mean of $x_{1}$ and $x_{2}$ is $M_{1}$ and that of $x_{1}, x_{2}, x_{3}, x_{4}$ is $M_{2}$ , then the mean of $a x_{1}, a x_{2}, \frac{x_{3}}{a}, \frac{x_{4}}{a}$ is