If the values 1 1-2- 1-3- 1-4- 1-5- -1-n occur at frequencies 1-2-3-4-5- - n in a distribution- then the mean isA- 1B- nC- 1-nD- 2-n-1

The correct option is D Mean = 1.1+1/2.2+1/3.3+1/4.4+1/5.5+...+1/nn/1+2+3+....+n =1+1+1+1+...+1/n(n+1)/2=n/n(n+1)/2=2/n+1. ...

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Byju's Answer

Standard IX

Mathematics

Mean

If the values...

Question

If the values $1 \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, . . . . . . \frac{1}{n}$ occur at frequencies $1, 2, 3, 4, 5, \dots \dots . .,$ n in a distribution, then the mean is

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Solution

The correct option is D
$M e a n = \frac{1.1 + \frac{1}{2} .2 + \frac{1}{3} .3 + \frac{1}{4} .4 + \frac{1}{5} .5 + . . . + \frac{1}{n} n}{1 + 2 + 3 + . . . . + n}$
$= \frac{1 + 1 + 1 + 1 + . . . + 1}{\frac{n (n + 1)}{2}} = \frac{n}{\frac{n (n + 1)}{2}} = \frac{2}{n + 1} .$

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If the values 112,13,14,15,......1n occur at frequencies 1,2,3,4,5,…….., n in a distribution, then the mean is

The correct option is D Mean=1.1+12.2+13.3+14.4+15.5+...+1nn1+2+3+....+n =1+1+1+1+...+1n(n+1)2=nn(n+1)2=2n+1.

If the values $1 \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, . . . . . . \frac{1}{n}$ occur at frequencies $1, 2, 3, 4, 5, \dots \dots . .,$ n in a distribution, then the mean is

The correct option is D
$M e a n = \frac{1.1 + \frac{1}{2} .2 + \frac{1}{3} .3 + \frac{1}{4} .4 + \frac{1}{5} .5 + . . . + \frac{1}{n} n}{1 + 2 + 3 + . . . . + n}$
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