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Let's say, I am sending network packets at the rate of 450 packets per second.

The size of these packets is uniformly distributed between 100 to 500 bytes.

I want to know what's the average packet size / per second?

I tried to solve it:

Since its uniformly distributed, so the average of size is (100 + 500)/2 = 300.

So, the answer is simply 300 bytes?

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  • $\begingroup$ I was surprised to find that the expected (=average) number of rolls of a $6$-sided die for rolling a certain value (say, 3 poins) at once is $6$. So yes, it's that simple. $\endgroup$ Commented Jun 1, 2020 at 0:59

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Let's add a bit of formalism here. Let $X$ (a random variable) be the size of the packet. We know that $X \text{~} \text{Unif}(100,500)$. Thus the expected size of a packet is $$\mathrm{E}(X)=\int_{100}^{500} {\frac{1}{500-100}x}\mathrm{d}x=300.$$ Then, the time it takes to send the average packet is $$\frac{300 \text{ bytes}}{450 \text{ bytes per second}}=2/3 \text{ second}$$

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  • $\begingroup$ Thanks, is it also correct if I say expected average size of packet = (100 + 500)/2 = 300 ? $\endgroup$
    – user963241
    Commented Jun 1, 2020 at 0:40
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    $\begingroup$ Your reasoning is completely valid as well. $\endgroup$
    – K.defaoite
    Commented Jun 1, 2020 at 0:40

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