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Let $S_{100}$ be the number of heads that turn up in 100 tosses of a fair coin. Use the Central Limit Theorem to estimate $P(S_{100} ≤ 45)$. How do I start off this question?

samir91
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Hint: You can write $S_{100}$ as the sum of $n=100$ i.i.d Bernoulli$(p=1/2)$ random variables $X_i$ i.e. $$S_{100}=\sum_{i=1}^{100}X_i$$ where $$X_i=\begin{cases}1, & \text{if Heads, } &p=\frac{1}{2}\\ 0, & \text{if Tails, }& q=1-p=\frac{1}{2}\end{cases}$$ Now, find $μ=E[X_i]$ and $σ^2=var(X_i)$ and use the CLT to obtain that $S_{100}$ is approximatelly normally distributed with parameters $μ_{100}=100μ$ and $σ_{100}^2=100σ^2$, in symbols $$S_{100}\sim N(100\cdot μ, 100 \cdot σ^2)$$ approximately.

Jimmy R.
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    Ok so I figured out the $\mu=1001/2$ since the probability of getting a head is $1/2$. Also, to get the standard variation, you do $1001/2*1/2$ Now we just follow the formula $P(S_n\le b) = P\left(Z\le\frac{b-n\mu}{\sqrt{n}\sigma}\right)$ Is this correct? – samir91 Dec 11 '14 at 21:47
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    Yes, but what you get like this is the variance, not the standard deviation. The rest is correct, see also here: http://math.stackexchange.com/questions/1058174/in-what-cases-do-i-use-these-formulas/1058279#1058279 – Jimmy R. Dec 11 '14 at 21:50
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    Yeah and the square root of the variance is the deviance so it all works out. Thanks. – samir91 Dec 11 '14 at 21:54