Consider any set of 201 observations x-1-x-2-x-200-x-201- It is given that x-1 lt- x-2 lt- - lt- x-200 lt-x-201- Then- the mean deviation of this set of observations about a point k is minimum- when k equals x-1 - x-2 - - - x-200-x-201-201-x-1-x-101-x-201-

The correct option is C x_101 x_1 lt; x_2 lt; x_3 lt; ⋯ lt; x_201 ∴ Median of the given observation = (201+1/2 )^th item =x_101 Now, deviations will ...

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Standard XII

Mathematics

Mean Deviation about Median

Consider any ...

Question

Consider any set of 201 observations $x_{1}, x_{2}, \dots, x_{200}, x_{201}$ . It is given that $x_{1} < x_{2} < \dots < x_{200} < x_{201}$ . Then, the mean deviation of this set of observations about a point k is minimum, when k equals

$\frac{x_{1} + x_{2} + \dots + x_{200} + x_{201}}{201}$

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$x_{1}$

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$x_{101}$

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$x_{201}$

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Solution

The correct option is C
$x_{101}$

$x_{1} < x_{2} < x_{3} < \dots < x_{201}$
$∴$ Median of the given observation $= {(\frac{201 + 1}{2})}^{th}$ item $= x_{101}$
Now, deviations will be minimum, if taken from the median.
$∴$ Mean deviation will be minimum, if k $= x_{101}$

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Similar questions

Consider any set of 201 observations $x_{1}, x_{2}, \dots, x_{200}, x_{201}$ . It is given that $x_{1} < x_{2} < \dots < x_{200} < x_{201}$ . Then, the mean deviation of this set of observations about a point k is minimum, when k equals

Q. Consider set of observations

x_{1}, x_{2}, x_{3}, \dots, x_{101}

. It is given that

x_{1} < x_{2} < x_{3} \dots < x_{100} < x_{101}

then the mean deviation of the set of observations about a point

k

is minimum when

k

equals

Q. For the values

x_{1}, x_{2} . . . . . x_{101}

of a distribution

x_{1} < x_{2} < x_{3} < . . . . . < x_{100} < x_{101} .

The mean deviation of this distribution with respect to a number k will be minimum when k is equal to

Q. The mean square deviation of set of

n

observations

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

about a point

c

is defined as

\frac{1}{n} n \sum i = 1 {(x_{i} - c)}^{2}

The mean square deviation about

- 2

and

2

are

18

and

10

respectively, then standard deviation of this set of observations is

The mean square deviations of a set of observations $x_{1}, x_{2}, \dots, x_{n}$ about a point c is defined to be $\frac{1}{n} \sum_{i = 1}^{n} (x_{i} - c)^{2}$
The mean square deviations about -1 and +1 of a set of observations are 7 and 3, respectively. Find the standard deviation of this set of observations.

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Consider any set of 201 observations x1, x2, ⋯,x200, x201. It is given that x1<x2<⋯<x200<x201. Then, the mean deviation of this set of observations about a point k is minimum, when k equals

The correct option is C x101 x1<x2<x3<⋯<x201 ∴ Median of the given observation =(201+12)th item =x101 Now, deviations will be minimum, if taken from the median. ∴ Mean deviation will be minimum, if k =x101

Consider any set of 201 observations $x_{1}, x_{2}, \dots, x_{200}, x_{201}$ . It is given that $x_{1} < x_{2} < \dots < x_{200} < x_{201}$ . Then, the mean deviation of this set of observations about a point k is minimum, when k equals

The correct option is C
$x_{101}$

$x_{1} < x_{2} < x_{3} < \dots < x_{201}$
$∴$ Median of the given observation $= {(\frac{201 + 1}{2})}^{th}$ item $= x_{101}$
Now, deviations will be minimum, if taken from the median.
$∴$ Mean deviation will be minimum, if k $= x_{101}$