If n - 20 - E x - 5 - then - x is -

The correct option is A 1/4 n= 20 E(x)=5 σ _x= ? σ _x=E(x)/n = 5/20 σ _x=1/4 ...

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Monotonically Increasing Functions

If n = 20 ,...

Question

If $n = 20, E (x) = 5,$ then $σ_{x}$ is :

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n_{1} < n_{2} < n_{3} < n_{4} < n_{5}

n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20

(n_{1}, n_{2}, n_{3}, n_{4}, n_{5})

lim n \to \infty \frac{(1^{4} + 2^{4} + 3^{4} + . . . . + (2 n)^{4}) (1^{3 / 2} + 2^{3 / 2} + . . . . + n^{3 / 2})}{(1^{5} + 2^{5} + . . . . + n^{5}) (1^{1 / 2} + 2^{1 / 2} + . . . + n^{1 / 2})}

n_{1} < n_{2} < n_{3} < n_{4} < n_{5}

n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20.

(n_{1}, n_{2}, n_{3}, n_{4}, n_{5})

1 \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, . . . . . . \frac{1}{n}

1, 2, 3, 4, 5, \dots \dots . .,

{lim}_{n \to \infty} \frac{1^{4} + 2^{4} + 3^{4} + . . . + n^{4}}{n^{5}} - {lim}_{n \to \infty} \frac{1^{3} + 2^{3} + 3^{3} + . . . + n^{3}}{n^{5}}

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