# Normal distribution of the mean of a uniform distribution

I have $$\bar{X}$$ which is the mean of the numbers from the uniform distribution of $$[0, 1]$$ with $$n = 100$$.

I know that $$\mu = \frac{1}{2}$$ and $$\sigma^2 = \frac{1}{1200}$$, thus, $$\sigma \approx 0.028$$, I need to find the probability of $$\bar{X}$$ having a value between $$[0.47, 0.53]$$.

Calculating the normal distribution, I find $$1 - 2(0.3508) = 0.2984$$, however the book says the answer is $$0.7016$$.

It is easy to see that $$1 - 0.2984 = 0.7016$$, and it makes me sure that I'm in the right path. I just don't know why I should subtract the value I found in the distribution from $$1$$, and it makes me think the book forgot one step before finishing the exercise. Could someone clarify this for me?

• Wolfram seems to agree with your book. wolframalpha.com/input/… Nov 28 '21 at 16:46
• By the CLT you can assume $\bar X \stackrel{aprx}{\sim}\mathsf{Norm}(\mu=.5,\sigma=\sqrt{1/1200})$ Then R code > diff(pnorm(c(.47,.53), .5, sqrt(1/1200))) returns $0.7013024.$ // If the textbook answer is based on printed normal tables, the slight discrepancy is due to rounding error using the tables. Nov 28 '21 at 18:07
• Also, by simulation in R with a million means of 100 uniform observations (to make sure 100 is plenty to use CLT). Code set.seed(2021); avg = replicate(10^6, mean(runif(100))); mean(avg> .47 & avg < .53) returns $0.700616$ with aprx margin of simulation error from 2^sd(avg> .47 & avg < .53)/1000. gives $0.001373625.$ This implies answ is $0.7006\pm 0.0014.$ Nov 28 '21 at 18:23

Hint: $$P(-a.

These are the inner (orange) areas of the graph below.

For more detailed explanation see here.

In your case the standardized value is

$$a=\frac{0.53-0.5}{\frac1{\sqrt{1200}}}=0.03\cdot \sqrt{1200}=0.6\cdot \sqrt{3}=1.03923...\approx 1.04$$

This table gives $$\Phi(a)=\Phi(1.04)=0.85083\approx 0.8508$$

Finally we get

$$P(-1.04

• Something wrong here. Will not downvote pending your revision. Nov 28 '21 at 18:09
• Can you give me a hint, Bruce? My final result is $0.7016$. Similar to yours. Nov 28 '21 at 18:12
• I missed that, maybe a final sentence would make it more obvious. Nov 28 '21 at 18:16
• Now I get it, I didn't pay attention to which area my table was referring, thanks Nov 28 '21 at 18:31
• @AkariOozora It pleases me that my answer met your requirements. Nov 28 '21 at 18:36