91Ó°ÊÓ

$X1,X2,...$be a sequence of independent and identically distributed random variables with the distribution$F$.

Using the central limit theorem, we have that

$\underset{i=1}{\overset{100}{∑}}{X}_i~AN\left(100\mathrm{μ},{\mathrm{σ}}^2\right)$

where$\mathrm{μ}=E\left(X_1\right)$and${\mathrm{σ}}^2=Var\left(X_1\right)$. Now suppose that$X_i~Pois\left(\mathrm{λ}\right)$. Therefore$\mathrm{μ}={\mathrm{σ}}^2=\mathrm{λ}$. We know that the sum of independent Poisson distributions is Poisson distribution. Therefore

$\underset{i=1}{\overset{100}{∑}}{X}_i~Pois\left(100\mathrm{λ}\right)$

which, implies that$E\left(\underset{i=1}{\overset{100}{∑}}{X}_i\right)=Var\left(\underset{i=1}{\overset{100}{∑}}{X}_i\right)=100\mathrm{λ}$. On the other hand,$CLT$gives us an approximation

$\underset{i=1}{\overset{100}{∑}}{X}_i~AN\left(100\mathrm{μ},{\mathrm{σ}}^2\right)=AN(100\mathrm{λ},\mathrm{λ})$

so the approximations for mean and variance arerole="math" localid="1649874336180" $E\left(\underset{i=1}{\overset{100}{∑}}{X}_i\right)≈100\mathrm{λ}$and$Var\left(\underset{i=1}{\overset{100}{∑}}{X}_i\right)=\mathrm{λ}$. Even though the approximation for the mean is exact, the approximation for the variance is significantly wrong since$100\mathrm{λ}≉\mathrm{λ}$. Therefore, we can conclude that the normal approximation $\underset{i=1}{\overset{100}{∑}}{X}_i$is not good.