The mean of $994$ , $996$ , $998$ , $1000$ and $1002$ is

992

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1004

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998

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999

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Solution

The correct option is C $998$
Given observations: $994, 996, 998, 1000, 1002$
Mean $= \frac{S u m}{N u m b e r o f o b s e r v a t i o n s}$

Mean $= \frac{994 + 996 + 998 + 1000 + 1002}{5}$

Mean $= \frac{4990}{5}$

Mean $= 998$

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Similar questions

996 \times 998 = ?

p

x^{50}

(1 + x)^{1000} + 2 x (1 + x)^{999} + 3 x^{2} (1 + x)^{998} + . . . . + 1001 x^{1000}

^{1002} C_{p}

(998 \times 1002)

x^{50}

(1 + x)^{1000} + 2 x (1 + x)^{999} + 3 x^{2} (1 + x)^{998} + . . . . . . + 1001 x^{1000}

x^{50}

(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + . . . + x^{1000}

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