$X_1,X_2 \dots X_n$ are independent R.V(random vars) that are uniform $\in$ [0, 1]

and let $S_n = X_1 + \dots + X_n$.

Now, I am trying to use the Central Limit Theorem to give an approximation of

P$(S_{100} \in [40, 60]).$

This seems like a fairly straightforward question. But I'm not sure how to approach it.

This is my attempt so far:

Since $X_i$ are uniformly distributed, they are uniform RV.

then E[X] = $\frac{40 + 60}{2}$ = 50 and my Variance = $\frac{(60-40)^2}{12} = 33.333$

I think I'm lost. Would appreciate any help and guidance! thanks


2 Answers 2


While $X$ has a uniform distribution, $S$ does not. Therefore you cannot use directly the mean and variance formulas that apply for a uniform distribution to find $E(S)$ and $V(S),$ as you have tried to do.

Here is an outline of what you need to do.

By the formulas for a uniform distribution, you have $E(X_i) = 1/2$ and $V(X_i) = 1/12.$ To find $E(S)$ you proceed as follows. using a rule that says "expectation of sum is sum of expectations":

$$E(S_{100}) = E(X_1 + X_2 + \cdots + X_{100}) = \sum_{i=1}^{100} E(X_i) = 100(1/2) = 50.$$

Because the $X_i$ are independent, you can use a similar method to add the 100 variances to get $V(S_{100}) = 100(1/12) = 100/12 = 8.3333.$ Then $SD(S_{100}) = \sqrt{100/12} = 2.887.$

Now, by the Central Limit Theorem, $S_{100} \approx W,$ where $W \sim \mathrm{Norm}(\mu = 50, \sigma = 2.887).$ So your problem becomes the evaluation of $P(40 \le W \le 60) \approx P(40 \le S_{100} \le 60).$ I assume you know how to standardize $W$ and use standard normal tables to evaluate $P(40 \le W \le 60)$.

Using software I get an answer that is very nearly 1. This makes sense because a normal distribution has almost all of its area within three standard deviations of the mean. This would be the interval $50 \pm 3(2.887)$ or roughly $(41.3, 58.7),$ which is contained in $(40, 60).$

  • $\begingroup$ Thank you for the clarification! $\endgroup$
    – misheekoh
    Mar 21, 2016 at 19:22
  • $\begingroup$ The second sentence here is nonsense, and is contradicted by the third paragraph . $\endgroup$ May 27, 2020 at 11:47
  • 1
    $\begingroup$ You have to read the question to understand the context of the second sentence. I will rewrite it. $\endgroup$
    – BruceET
    May 27, 2020 at 17:37
  • $\begingroup$ +1 but I think it would be better to say "by the Central Limit Theorem $\frac{1}{\sqrt{n}} (S_{100} - E[S_{100}]) \approx W$ where $W$ is normal with mean 0 and variance $Var(X_1)$." The issue with approximating the non-scaled $S_{100}$ as Gaussian becomes clearer if $X_i$ were Bernoulli, then $S_{100}$ would always be an integer and not approximately Gaussian, though its scaled version can be viewed as approximately Gaussian in distribution. $\endgroup$
    – Michael
    May 27, 2020 at 19:25
  • 2
    $\begingroup$ Please feel free to give your own Answer. $\endgroup$
    – BruceET
    May 28, 2020 at 1:57

Let's start by looking at the approximate distribution of $S_n$ when $n$ is large, by applying the CLT. Since you have underlying uniform values $X_1,X_2,X_3,... \sim \text{IID U}(0,1)$, you have $\mathbb{E}(X_i) = \tfrac{1}{2}$ and $\mathbb{V}(X_i) = \tfrac{1}{12}$. The sum of the first $n$ values has the corresponding moments:

$$\mathbb{E}(S_n) = \frac{n}{2} \quad \quad \quad \mathbb{V}(S_n) = \frac{n}{12}.$$

Thus, for large $n$ you can apply the CLT to get the approximate distribution $S_n \overset{\text{Approx}}{\sim} \text{N}( n/2, n/12 )$, or if you prefer to standardise you get:

$$\frac{2S_n - n}{\sqrt{n/3}} \overset{\text{Approx}}{\sim} \text{N} (0,1).$$

Thus, you have the general rule:

$$\begin{aligned} \mathbb{P}(s_* \leqslant S_n \leqslant s^*) &= \mathbb{P} \Big( \frac{2s_* - n}{\sqrt{n/3}} \leqslant \frac{2S_n - n}{\sqrt{n/3}} \leqslant \frac{2s^* - n}{\sqrt{n/3}} \Big) \\[6pt] &\approx \Phi \Big( \frac{2s^* - n}{\sqrt{n/3}} \Big) - \Phi \Big( \frac{2s_* - n}{\sqrt{n/3}} \Big). \\[6pt] \end{aligned}$$

For your particular values you have:

$$\begin{aligned} \mathbb{P}(40 \leqslant S_{100} \leqslant 60) &\approx \Phi \Big( \frac{2 \cdot 60 - 100}{\sqrt{100/3}} \Big) - \Phi \Big( \frac{2 \cdot 40 - 100}{\sqrt{100/3}} \Big) \\[6pt] &= \Phi \Big( \frac{20}{\sqrt{100/3}} \Big) - \Phi \Big( - \frac{20}{\sqrt{100/3}} \Big) \\[12pt] &= \Phi ( 2 \sqrt{3} ) - \Phi ( -2 \sqrt{3} ) \\[12pt] &= \Phi ( 3.464102 ) - \Phi ( -3.464102 ) \\[12pt] &= 0.999468. \\[12pt] \end{aligned}$$


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