If x- is the mean of x1- x2- - - xn- then the mean of ax1-ax2-axn-x1a-x2a-xna is-

The correct option is A ( a+1a )x̅2 Mean of x_1, x_2, ... , x_n=x̅ Mean of ax_1,ax_2,....ax_n,x_1a,x_2a,....,x_na #160; =ax_1 + ax_2 +... ...

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Standard IX

Mathematics

Average

If x̅ is th...

Question

If $¯ x$ is the mean of $x_{1}, x_{2}, . . ., x_{n}$ , then the mean of $a x_{1}, a x_{2}, . . . . a x_{n}, \frac{x_{1}}{a}, \frac{x_{2}}{a}, . . . ., \frac{x_{n}}{a}$ is:

(a + \frac{1}{a}) ¯ x

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(a + \frac{1}{a}) \frac{¯ x}{2}

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(a + \frac{1}{a}) \frac{¯ x}{n}

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\frac{(a + \frac{1}{a}) ¯ x}{2 n}

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Solution

The correct option is A $(a + \frac{1}{a}) \frac{¯ x}{2}$
Mean of $x_{1}, x_{2}, . . ., x_{n} = ¯ x$

Mean of $a x_{1}, a x_{2}, . . . . a x_{n}, \frac{x_{1}}{a}, \frac{x_{2}}{a}, . . . ., \frac{x_{n}}{a}$
$= \frac{a x_{1} + a x_{2} + . . . . a x_{n} + \frac{x_{1}}{a} + \frac{x_{2}}{a} + . . . . + \frac{x_{n}}{a}}{2 n}$

$= \frac{a (x_{1} + x_{2} + . . . + x_{n}) + \frac{1}{a} (x_{1} + x_{2} + . . . + x_{n})}{2 n}$

$= \frac{¯ x}{2} (a + \frac{1}{a})$

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Similar questions

Q. Question 16
If

¯ x

is the mean of

x_{1}, x_{2}, \dots \dots x_{n}

, then for

a \neq 0

, the mean of

a x_{1}, a x_{2}, \dots \dots a x_{n}, \frac{x_{1}}{a}, \frac{x_{2}}{a}, \dots \dots \frac{x_{n}}{a}

is:

A)

(a + \frac{1}{a}) ¯ x

(a + \frac{1}{a}) \frac{¯ x}{2}

(a + \frac{1}{a}) \frac{¯ x}{n}

\frac{(a + \frac{1}{a}) ¯ x}{2 n}

Q. Question 16
If

¯ x

is the mean of

x_{1}, x_{2}, \dots \dots x_{n}

, then for

a \neq 0

, the mean of

a x_{1}, a x_{2}, \dots \dots a x_{n}, \frac{x_{1}}{a}, \frac{x_{2}}{a}, \dots \dots \frac{x_{n}}{a}

is:

A)

(a + \frac{1}{a}) ¯ x

(a + \frac{1}{a}) \frac{¯ x}{2}

(a + \frac{1}{a}) \frac{¯ x}{n}

\frac{(a + \frac{1}{a}) ¯ x}{2 n}

Given that is the mean and σ² is the variance of n observations x₁, x₂ … x_n. Prove that the mean and variance of the observations ax₁, ax₂, ax₃ …ax_nare and a² σ², respectively (a ≠ 0).

Q. If the arithmetic mean of the number

x_{1}, x_{2}, x_{3} . . . . . . . . ., x_{n}

¯ ¯ ¯ x

, then the arithmetic mean of numbers

a x_{1}, + b, a x_{2} + b, a x_{3} + b, . . . . . . . . . . . . . a x_{n} + b,

where a, b are two constants would be

If $¯ x$ is the mean of $x_{1}, x_{2} \dots, x_{n}$ then for $a \neq 0$ , the mean of $a x_{1}, a x_{2} \dots, a x_{n}, \frac{x_{1}}{a}, \frac{x_{2}}{a}, \dots, \frac{x_{n}}{a}$ is
(a) $(a + \frac{1}{a}) ¯ x$
(a) $(a + \frac{1}{a}) \frac{¯ x}{2}$
(c) $(a + \frac{1}{a}) \frac{¯ x}{n}$
(d) $\frac{(a + \frac{1}{a}) ¯ x}{2 n}$

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If ¯x is the mean of x1,x2,...,xn, then the mean of ax1,ax2,....axn,x1a,x2a,....,xna is:

The correct option is A (a+1a)¯x2Mean of x1,x2,...,xn=¯xMean of ax1,ax2,....axn,x1a,x2a,....,xna=ax1+ax2+....axn+x1a+x2a+....+xna2n=a(x1+x2+...+xn)+1a(x1+x2+...+xn)2n=¯x2(a+1a)

If $¯ x$ is the mean of $x_{1}, x_{2}, . . ., x_{n}$ , then the mean of $a x_{1}, a x_{2}, . . . . a x_{n}, \frac{x_{1}}{a}, \frac{x_{2}}{a}, . . . ., \frac{x_{n}}{a}$ is: