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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?

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  • $\begingroup$ Wolfram seems to agree with your book. wolframalpha.com/input/… $\endgroup$
    – Blitzer
    Nov 28 '21 at 16:46
  • $\begingroup$ 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. $\endgroup$
    – BruceET
    Nov 28 '21 at 18:07
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    $\begingroup$ 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 $0.7006\pm 0.0014.$ $\endgroup$
    – BruceET
    Nov 28 '21 at 18:23
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Hint: $P(-a<X<a)=2\Phi(a)-1$.

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

For more detailed explanation see here.

Additional:

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

$$P(-1.04<X<1.04)=2\Phi(1.04)-1=2\cdot 0.8508-1=1.7016-1=0.7016$$

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

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