 Yasmin

2021-09-03

Let ${X}_{1},{X}_{2},\stackrel{˙}{s},{X}_{n}$ be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables. Find n such that Pr$\left(X>2000\right)\ge 0.95$ Faiza Fuller

Expert

Step 1
${X}_{1},{X}_{2},\stackrel{˙}{s},{X}_{n}$ are independent ranodom variables with mean 100 and standard deviation 30
So, Var$\left({X}_{i}\right)={30}^{2}=900$
Now given that,
$X=\sum _{i}{X}_{i}$
So $E\left(X\right)=E\left(\sum _{i}{X}_{i}\right)=\sum _{i}E\left({X}_{i}\right)=100n$
and $Var\left(X\right)=Var\left(\sum _{i}{X}_{i}\right)=\sum _{i}Var\left({X}_{i}\right)$ [ Covariances are zero due to independence]
So, Var(X) = 900n
Step 2
By Central Limit Theorem,
$\frac{X-\mu }{\sigma }\equiv \frac{X-100n}{\sqrt{900n}}\sim N\left(0,1\right)$
Now, P(X>2000)>0,95
implies $1-P\left(X\le 2000\right)>0,95$
$P\left(X\le 2000\right)<0,05$
$P\left(\frac{X-100n}{\sqrt{900n}}\le \frac{2000-100n}{30\sqrt{n}}\right)<0,05$
implies $\mathrm{\Phi }\left[\frac{2000-100n}{30\sqrt{n}}\right]<0,05=\mathrm{\Phi }\left(-1,645\right)$
implies $\frac{2000-100n}{30\sqrt{n}}<-1,645$
Here you can solve this in two methods,
Trial and error method
Numerically using root finding method
Using trial and error method,
For n = 22
$\frac{2000-100n}{30\sqrt{n}}=-1,4213$
For n=23
$\frac{2000-100n}{30\sqrt{n}}=-2,085$
So, for n = 23 it is crossing -1.645. Hence n = 23

Do you have a similar question?