PROBABILITY
Stirling’s approximation for largexgives
x!≈√
2 πx(xe)ximplying that
lnx!≈ln√
2 πx+xlnx−x,which, on substituting into (30.115), yields
lnf(x)≈−λ+xlnλ−(xlnx−x)−ln√
2 πx.Since we expect the Poisson distribution to peak aroundx=λ, we substitute
=x−λto obtain
lnf(x)≈−λ+(λ+){
lnλ−ln[
λ(
1+
λ)]}
+(λ+)−ln√
2 π(λ+).Using the expansion ln(1 +z)=z−z^2 /2+···, we find
lnf(x)≈−(λ+)(
λ−^2
2 λ^2)
−ln√
2 πλ−(
λ−^2
2 λ^2)≈−^2
2 λ−ln√
2 πλ,when only the dominant terms are retained, after using the fact thatis of the
order of the standard deviation ofx,i.e.oforderλ^1 /^2. On exponentiating this
result we obtain
f(x)≈1
√
2 πλexp[
−(x−λ)^2
2 λ]
,which is the Gaussian distribution withμ=λandσ^2 =λ.
The larger the value ofλ, the better is the Gaussian approximation to thePoisson distribution; the approximation is reasonable even forλ= 5, butλ≥ 10
is safer. As in the case of the Gaussian approximation to the binomial distribution,
a continuity correction is necessary since the Poisson distribution is discrete.
E-mail messages are received by an author at an average rate of one per hour. Find the
probability that in a day the author receives 24 messages or more.We first define the random variable
X= number of messages received in a day.ThusE[X]=1×24 = 24, and soX∼Po(24). Sinceλ>10 we may approximate the
Poisson distribution byX∼N(24, 24). Now the standard variable is
Z=X− 24
√
24
,
and, using the continuity correction, we find
Pr(X> 23 .5) = Pr(
Z>
23. 5 − 24
√
24
)
=Pr(Z>− 0 .102) = Pr(Z< 0 .102) = 0. 54 .