Editing Binomial distribution (section)

=== Normal approximation ===
{{see also|Binomial proportion confidence interval#Normal approximation interval}}

[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for {{math|1=''n'' = 6}} and {{math|1=''p'' = 0.5}}]]

If {{math|''n''}} is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to {{math|B(''n'', ''p'')}} is given by the [[normal distribution]]
: <math> \mathcal{N}(np,\,np(1-p)),</math>
and this basic approximation can be improved in a simple way by using a suitable [[continuity correction]].
The basic approximation generally improves as {{math|''n''}} increases (at least 20) and is better when {{math|''p''}} is not near to 0 or 1.<ref name="bhh">{{cite book|title=Statistics for experimenters|url=https://archive.org/details/statisticsforexp00geor|url-access=registration|author=Box, Hunter and Hunter|publisher=Wiley|year=1978|page=[https://archive.org/details/statisticsforexp00geor/page/130 130]|isbn=9780471093152}}</ref> Various [[Rule of thumb|rules of thumb]] may be used to decide whether {{math|''n''}} is large enough, and {{math|''p''}} is far enough from the extremes of zero or one:
* One rule<ref name="bhh"/> is that for {{math|''n'' > 5}} the normal approximation is adequate if the absolute value of the skewness is strictly less than 0.3; that is, if
*: <math>\frac{|1-2p|}{\sqrt{np(1-p)}}=\frac1{\sqrt{n}}\left|\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}\,\right|<0.3.</math>
This can be made precise using the [[Berry–Esseen theorem]].
* A stronger rule states that the normal approximation is appropriate only if everything within 3 standard deviations of its mean is within the range of possible values; that is, only if
*: <math>\mu\pm3\sigma=np\pm3\sqrt{np(1-p)}\in(0,n).</math>
: This 3-standard-deviation rule is equivalent to the following conditions, which also imply the first rule above.
:: <math>n>9 \left(\frac{1-p}{p} \right)\quad\text{and}\quad n>9\left(\frac{p}{1-p}\right).</math>
{{hidden begin|style=width:66%|ta1=center|border=1px #aaa solid|title=[Proof]}}
The rule <math> np\pm3\sqrt{np(1-p)}\in(0,n)</math> is totally equivalent to request that
: <math>np-3\sqrt{np(1-p)}>0\quad\text{and}\quad np+3\sqrt{np(1-p)}<n.</math>
Moving terms around yields:
: <math>np>3\sqrt{np(1-p)}\quad\text{and}\quad n(1-p)>3\sqrt{np(1-p)}.</math>
Since <math>0<p<1</math>, we can apply the square power and divide by the respective factors <math>np^2</math> and <math>n(1-p)^2</math>, to obtain the desired conditions:
: <math>n>9 \left(\frac{1-p}p\right) \quad\text{and}\quad n>9 \left(\frac{p}{1-p}\right).</math>
Notice that these conditions automatically imply that <math>n>9</math>. On the other hand, apply again the square root and divide by 3,
: <math>\frac{\sqrt{n}}3>\sqrt{\frac{1-p}p}>0 \quad \text{and} \quad \frac{\sqrt{n}}3 > \sqrt{\frac{p}{1-p}}>0.</math>
Subtracting the second set of inequalities from the first one yields:
: <math>\frac{\sqrt{n}}3>\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}>-\frac{\sqrt{n}}3;</math>
and so, the desired first rule is satisfied,
: <math>\left|\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}\,\right|<\frac{\sqrt{n}}3.</math>
{{hidden end}}
* Another commonly used rule is that both values {{math|''np''}} and {{math|''n''(1 − ''p'')}} must be greater than<ref>{{Cite book |last=Chen |first=Zac |title=H2 Mathematics Handbook |publisher=Educational Publishing House |year=2011 |isbn=9789814288484 |edition=1 |location=Singapore |pages=350}}</ref><ref>{{Cite web |date=2023-05-29 |title=6.4: Normal Approximation to the Binomial Distribution - Statistics LibreTexts |url=https://stats.libretexts.org/Courses/Las_Positas_College/Math_40:_Statistics_and_Probability/06:_Continuous_Random_Variables_and_the_Normal_Distribution/6.04:_Normal_Approximation_to_the_Binomial_Distribution |access-date=2023-10-07 |archive-date=2023-05-29 |archive-url=https://web.archive.org/web/20230529211919/https://stats.libretexts.org/Courses/Las_Positas_College/Math_40:_Statistics_and_Probability/06:_Continuous_Random_Variables_and_the_Normal_Distribution/6.04:_Normal_Approximation_to_the_Binomial_Distribution |url-status=bot: unknown }}</ref> or equal to&nbsp;5. However, the specific number varies from source to source, and depends on how good an approximation one wants. In particular, if one uses&nbsp;9 instead of&nbsp;5, the rule implies the results stated in the previous paragraphs.
{{hidden begin|style=width:66%|ta1=center|border=1px #aaa solid|title=[Proof]}}
Assume that both values <math>np</math> and <math>n(1-p)</math> are greater than&nbsp;9. Since <math>0< p<1</math>, we easily have that 
: <math>np\geq9>9(1-p)\quad\text{and}\quad n(1-p)\geq9>9p.</math>
We only have to divide now by the respective factors <math>p</math> and <math>1-p</math>, to deduce the alternative form of the 3-standard-deviation rule:
: <math>n>9 \left(\frac{1-p}p\right) \quad\text{and}\quad n>9 \left(\frac{p}{1-p}\right).</math>
{{hidden end}}

The following is an example of applying a [[continuity correction]]. Suppose one wishes to calculate {{math|Pr(''X'' ≤ 8)}} for a binomial random variable {{math|''X''}}. If {{math|''Y''}} has a distribution given by the normal approximation, then {{math|Pr(''X'' ≤ 8)}} is approximated by {{math|Pr(''Y'' ≤ 8.5)}}. The addition of 0.5 is the continuity correction; the uncorrected normal approximation gives considerably less accurate results.

This approximation, known as [[de Moivre–Laplace theorem]], is a huge time-saver when undertaking calculations by hand (exact calculations with large {{math|''n''}} are very onerous); historically, it was the first use of the normal distribution, introduced in [[Abraham de Moivre]]'s book ''[[The Doctrine of Chances]]'' in 1738. Nowadays, it can be seen as a consequence of the [[central limit theorem]] since {{math|B(''n'', ''p'')}} is a sum of {{math|''n''}} independent, identically distributed [[Bernoulli distribution|Bernoulli variables]] with parameter&nbsp;{{math|''p''}}. This fact is the basis of a [[hypothesis test]], a "proportion z-test", for the value of {{math|''p''}} using {{math|''x''/''n''}}, the sample proportion and estimator of {{math|''p''}}, in a [[common test statistics|common test statistic]].<ref>[[NIST]]/[[SEMATECH]], [http://www.itl.nist.gov/div898/handbook/prc/section2/prc24.htm "7.2.4. Does the proportion of defectives meet requirements?"] ''e-Handbook of Statistical Methods.''</ref>

For example, suppose one randomly samples {{math|''n''}} people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of ''n'' people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion ''p'' of agreement in the population and with standard deviation
: <math>\sigma = \sqrt{\frac{p(1-p)}{n}}</math>