For the frequency distribution : Variate (x) : x1 x2 x3 .... x15 Frequency (f) : f1 f2 f3 ...... f15 where 0 0, the standard deviation cannot be :

Statistics PYQ — JEE Main 2020 · 3 Sep · Shift 1 · Q19 with Solution

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Correct answer: A

Let the mean of the distribution be $\bar x=\frac{\sum f_i x_i}{\sum f_i}$ and the standard deviation be $\sigma=\sqrt{\frac{\sum f_i(x_i-\bar x)^2}{\sum f_i}}.$
Since there are $15$ distinct variates $x_1<x_2<\cdots<x_{15}$ and frequencies $f_i>0$ , every one of these $15$ values is present in the distribution.
For any dataset whose minimum value is $a$ and maximum value is $b$ , the standard deviation satisfies $\sigma\le \frac{b-a}{2}.$ This is because all observations lie in the interval $[a,b]$ .
Here there are $15$ distinct values, so between the smallest and largest values there must be at least $14$ gaps of size at least $1$ (since the variates are distinct, typically taken as ordered real values; for JEE statistics questions this implies the least possible spread for $15$ distinct variates is attained by consecutive values). Hence $x_{15}-x_1\ge 14.$ So the maximum possible standard deviation is at least not unrestricted. But we need a sharper fact.
The largest possible standard deviation for $15$ distinct values in a span of $14$ occurs when the values are equally spread as $a,a+1,a+2,\dots,a+14$ with equal frequencies. Then $\bar x=a+7$ and

=\frac{2(1^2+2^2+\cdots+7^2)}{15}.$$ Now, $$1^2+2^2+\cdots+7^2=\frac{7\cdot 8\cdot 15}{6}=140.$$ Therefore, $$\sigma^2=\frac{2\cdot 140}{15}=\frac{280}{15}=\frac{56}{3},$$ so $$\sigma=\sqrt{\frac{56}{3}}\approx 4.32.$$ Thus standard deviation cannot be $6$. 6. The other values are possible in suitable distributions: - $\sigma=1$, possible. - $\sigma=2$, possible. - $\sigma=4$, possible since it is below the attainable upper range. Hence the value that standard deviation cannot be is $$\boxed{6}.$$

View written solutionFree

Correct answer: A

Let the mean of the distribution be $\bar x=\frac{\sum f_i x_i}{\sum f_i}$ and the standard deviation be $\sigma=\sqrt{\frac{\sum f_i(x_i-\bar x)^2}{\sum f_i}}.$
Since there are $15$ distinct variates $x_1<x_2<\cdots<x_{15}$ and frequencies $f_i>0$ , every one of these $15$ values is present in the distribution.
For any dataset whose minimum value is $a$ and maximum value is $b$ , the standard deviation satisfies $\sigma\le \frac{b-a}{2}.$ This is because all observations lie in the interval $[a,b]$ .
Here there are $15$ distinct values, so between the smallest and largest values there must be at least $14$ gaps of size at least $1$ (since the variates are distinct, typically taken as ordered real values; for JEE statistics questions this implies the least possible spread for $15$ distinct variates is attained by consecutive values). Hence $x_{15}-x_1\ge 14.$ So the maximum possible standard deviation is at least not unrestricted. But we need a sharper fact.
The largest possible standard deviation for $15$ distinct values in a span of $14$ occurs when the values are equally spread as $a,a+1,a+2,\dots,a+14$ with equal frequencies. Then $\bar x=a+7$ and

=\frac{2(1^2+2^2+\cdots+7^2)}{15}.$$ Now, $$1^2+2^2+\cdots+7^2=\frac{7\cdot 8\cdot 15}{6}=140.$$ Therefore, $$\sigma^2=\frac{2\cdot 140}{15}=\frac{280}{15}=\frac{56}{3},$$ so $$\sigma=\sqrt{\frac{56}{3}}\approx 4.32.$$ Thus standard deviation cannot be $6$. 6. The other values are possible in suitable distributions: - $\sigma=1$, possible. - $\sigma=2$, possible. - $\sigma=4$, possible since it is below the attainable upper range. Hence the value that standard deviation cannot be is $$\boxed{6}.$$