JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 1)
For the frequency distribution :
Variate (x) : x1 x2 x3 .... x15
Frequency (f) : f1 f2 f3 ...... f15
where 0 < x1 < x2 < x3 < ... < x15 = 10 and
$$\sum\limits_{i = 1}^{15} {{f_i}} $$ > 0, the standard deviation cannot be :
Variate (x) : x1 x2 x3 .... x15
Frequency (f) : f1 f2 f3 ...... f15
where 0 < x1 < x2 < x3 < ... < x15 = 10 and
$$\sum\limits_{i = 1}^{15} {{f_i}} $$ > 0, the standard deviation cannot be :
6
1
4
2
Explanation
If variate varries from m to M then variance
$${\sigma ^2} \le {1 \over 4}{\left( {M - m} \right)^2}$$
(M = upper bound of value of any random variable,
m = Lower bound of value of any random variable)
Here M = 10 and m = 0
$$ \therefore $$ $${\sigma ^2} \le {1 \over 4}{\left( {10 - 0} \right)^2}$$
$$ \Rightarrow $$ $${\sigma ^2} \le 25$$
$$ \Rightarrow $$ $$ - 5 \le \sigma \le 5$$
$$ \therefore $$ $$\sigma $$ $$ \ne $$ 6
$${\sigma ^2} \le {1 \over 4}{\left( {M - m} \right)^2}$$
(M = upper bound of value of any random variable,
m = Lower bound of value of any random variable)
Here M = 10 and m = 0
$$ \therefore $$ $${\sigma ^2} \le {1 \over 4}{\left( {10 - 0} \right)^2}$$
$$ \Rightarrow $$ $${\sigma ^2} \le 25$$
$$ \Rightarrow $$ $$ - 5 \le \sigma \le 5$$
$$ \therefore $$ $$\sigma $$ $$ \ne $$ 6
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