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In a specific design, the true voltage of a circuit is 250 millivolt(mV). Measurement error that is continuous and uniformly distributed from -3 to +3 mV is added to the true voltage. the measurement is then rounded to the nearest mV so that it becomes discrete.

(i) Write down the probability mass function of the measured voltage.

(ii) What is the mean and variance of the measured voltage?

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    $\begingroup$ Welcome to the site! What have you tried out? Showing some attempt might increase your chances of getting an answer. $\endgroup$ – Easymode44 Mar 8 at 14:01
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If we take the continuous uniform distribution over $[257,263]$ we see that the PDF will be $\frac{1}{6}, V\in[257,263]$.

Now if we round it, notice that the interval $[257.5,258.5)$ with a range of 1 will be rounded to $268$ but $[257,257.5)$ with a range of 1/2 will be rounded to $267$.

So our PMF looks like this

$$\begin{array}{r|ccccccc} V & 257 & 258 & 259 & 260 & 261 & 262 & 263 \\ \hline \mathrm{P}(V=v) & \dfrac{1}{12} & \dfrac{1}{6} & \dfrac{1}{6} & \dfrac{1}{6} & \dfrac{1}{6} & \dfrac{1}{6} & \dfrac{1}{12} \end{array}$$

Or in other words

$$\mathrm{P}(V=v)=\begin{cases} 1/6 & V=258,259,260,261,262 \\ 1/12 & V=257,263 \end{cases}$$

And now you can use the formula for the mean and variance

$$ \mathrm{E}(V) = \sum v\,\mathrm{P}(V=v) \\ \mathrm{Var}(V) = \mathrm{E}(V^2) - \mathrm{E}(V)^2$$

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