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