# Let X_1,X_2 , dots , X_n be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables.

Let $$\displaystyle{X}_{{1}},{X}_{{2}},\dot{{s}},{X}_{{n}}$$ 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$$\displaystyle{\left({X}{>}{2000}\right)}\geq{0.95}$$

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Step 1
$$\displaystyle{X}_{{1}},{X}_{{2}},\dot{{s}},{X}_{{n}}$$ are independent ranodom variables with mean 100 and standard deviation 30
So, Var$$\displaystyle{\left({X}_{{i}}\right)}={30}^{{2}}={900}$$
Now given that,
$$\displaystyle{X}=\sum_{{i}}{X}_{{i}}$$
So $$\displaystyle{E}{\left({X}\right)}={E}{\left(\sum_{{i}}{X}_{{i}}\right)}=\sum_{{i}}{E}{\left({X}_{{i}}\right)}={100}{n}$$
and $$\displaystyle{V}{a}{r}{\left({X}\right)}={V}{a}{r}{\left(\sum_{{i}}{X}_{{i}}\right)}=\sum_{{i}}{V}{a}{r}{\left({X}_{{i}}\right)}$$ [ Covariances are zero due to independence]
So, Var(X) = 900n
Step 2
By Central Limit Theorem,
$$\displaystyle{\frac{{{X}-\mu}}{{\sigma}}}\equiv{\frac{{{X}-{100}{n}}}{{\sqrt{{{900}{n}}}}}}\sim{N}{\left({0},{1}\right)}$$
Now, P(X>2000)>0,95
implies $$\displaystyle{1}-{P}{\left({X}\leq{2000}\right)}{>}{0},{95}$$
$$\displaystyle{P}{\left({X}\leq{2000}\right)}{<}{0},{05}$$
$$\displaystyle{P}{\left({\frac{{{X}-{100}{n}}}{{\sqrt{{{900}{n}}}}}}\leq{\frac{{{2000}-{100}{n}}}{{{30}\sqrt{{n}}}}}\right)}{<}{0},{05}$$
implies $$\displaystyle\Phi{\left[{\frac{{{2000}-{100}{n}}}{{{30}\sqrt{{n}}}}}\right]}{<}{0},{05}=\Phi{\left(-{1},{645}\right)}$$
implies $$\displaystyle{\frac{{{2000}-{100}{n}}}{{{30}\sqrt{{n}}}}}{<}-{1},{645}$$
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
$$\displaystyle{\frac{{{2000}-{100}{n}}}{{{30}\sqrt{{n}}}}}=-{1},{4213}$$
For n=23
$$\displaystyle{\frac{{{2000}-{100}{n}}}{{{30}\sqrt{{n}}}}}=-{2},{085}$$
So, for n = 23 it is crossing -1.645. Hence n = 23