If the mean deviation about the mean of the numbers 1 , 2 , 3 , … … , n , where n is odd, is 5 n + 1 n , then n is equal to ______.

We know that mean deviation about the mean of odd natural number is given by M.D x = 1 n n 2 - 1 4 So, now 1 n n 2 - 1 4 = 5 n + 1 n (given) ⇒ n - 1 = 20 ⇒ n = 21

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MathematicsStatisticsMeasures of DispersionMedium2 minPYQ_2022

MathematicsMediumnumerical

If the mean deviation about the mean of the numbers $1, 2, 3, \dots \dots, n$ , where $n$ is odd, is $\frac{5 (n + 1)}{n}$ , then $n$ is equal to ______.

Answer:

21.00

Solution:

We know that mean deviation about the mean of odd natural number is given by $M.D (x) = \frac{1}{n} [\frac{n^{2} - 1}{4}]$

So, now $\frac{1}{n} [\frac{n^{2} - 1}{4}] = \frac{5 (n + 1)}{n}$ (given)

$\Rightarrow n - 1 = 20 \Rightarrow n = 21$

Stream:JEESubject:MathematicsTopic:StatisticsSubtopic:Measures of Dispersion

⏱ 2mℹ️ Source: PYQ_2022

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