For the frequency distribution: Variate (x): $\quad \mathrm{x}_{1} \quad \mathrm{x}_{2} \quad \mathrm{x}_{3} \ldots . \mathrm{x}_{15}$ Frequency (f): $f_{1} \quad f_{2} \quad f_{3} \ldots . f_{15}$ Where $0 0$, the standard deviation cannot be:

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Statistics

For the frequency distribution:

Variate (x):

x_{1} x_{2} x_{3} \dots . x_{15}

Frequency (f):

f_{1} f_{2} f_{3} \dots . f_{15}

Where

0 < x_{1} < x_{2} < x_{3} < \dots < x_{15} = 10

and

\sum_{i = 1}^{15} f_{i} > 0

, the standard deviation cannot be:

Answer: (C)

For data confined to

[m, M]

the variance can be at most

(\frac{M - m}{2})^{2}

. Here the values run from just above

0

up to

10

, so

σ^{2} < (\frac{10}{2})^{2} = 25

and

σ < 5

. A standard deviation of

6

is therefore impossible.

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option