# uniformly distributed question…

I am doing self-study. but I don`t calculate this problem. so I want to calculating course.

Three numbers are randomly selected and rounded-off to the nearest interger. Let $X_1, X_2, X_3$ denote the round-off errors and assume that they are independent and uniformly distributed over $[-0.4,0.4]$. Let $D$ be the sum of the errors: $D=X_1+X_2+X_3$

1) Compute $E(D)$

2) Compute $\mathrm{Var}(D)$

3) Use the Central Limit Theorem to approximate the probability $P(D>0.1)$. Give the answer in terms of the cdf $\Phi(z)=P(Z<-z)$ of $Z\sim N(0,1)$.

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We first calculate the mean and variance of $X_1$. By symmetry, the mean is $0$. If roundoff errors are indeed uniformly distributed on $[-0.4,0.4]$, then the density function of $X_1$ is $\dfrac{1}{0.8}$ over this interval, and $0$ elsewhere. The variance of $X_1$ is $E(X_1^2)-(E(X_1))^2$. We have $$E(X_1^2)=\int_{-0.4}^{0.4}\frac{1}{0.8}x^2\,dx.$$ Integrate. We get $\dfrac{(0.4)^2}{3}$.
Thus $X_1$ has variance $\dfrac{(0.4)^2}{3}$. So do $X_2$ and $X_3$. (You may have been given a formula for the variance of a uniform on $[a,b]$. In that case, you could just use that formula.)
Since $D=X_1+X_2+X_3$, we see that $D$ has mean $0$ and variance $(0.4)^2$, and therefore standard deviation $0.4$.
Now you are asked to use CLT to approximate the probability that $D\gt 0.1$. This is a straighforward normal distribution mean $0$ standard deviation $0.4$ calculation. I am a bit uncomfortable with using CLT, since $3$ is a very small sample size. The answer, if we use CLT, turns out to be $1-\Pr\left(Z\le \dfrac{0.1}{0.4}\right)$.