Central Limit Theorem. How to apply to the task. The research showed that the probabilities of 3, 4, 5, 6 and 7 cars broken on one day are 0.3, 0.4, 0.2, 0.08, 0.02 respectively. If 221 car broke in 50 days, does it show that more cars break than expected in the research. 
I know I can use CLT in here. 
I calculated the expected value - its 4.12 for a car to break. The variance is 0.986. Wekepedia suggests that sqrt(nnumber)(average-Expected) tends to be close to normal distribution as more random variables are given. 
But how I cant tie up the CLT and the proof. How can this task be solved?
 A: This is a nice problem, and I would like to share its exact solution (that does not use CLT) using help of Mathematica 9:
Start by defining the distribution, as determined by research:
probs = {0.3, 0.4, 0.2, 0.08, 0.02};
cars = Range[3, 7];

di = EmpiricalDistribution[WeightedData[cars, probs]];

Next, find the weights of the distribution for $Y=X_1+\cdots+X_{50}$ using convolution:
di50 = With[{m = 50}, 
   EmpiricalDistribution[
    Nest[ListConvolve[probs, ArrayPad[#, Length[probs] {1, 1} - 1]] &,
       probs, m - 1] -> Range[Min[cars] m, Max[cars] m]]];

The plot of the probability mass function $p_Y(y)$ resembles normal distribution, as CLT tells us:

We can now ask for probability that $Y$ deviates from its mean by the greater amount that the current measurement of 221 does:
In[76]:= NProbability[Abs[Y - Mean[di50]] >= Abs[221 - Mean[di50]], 
 Y \[Distributed] di50]

Out[76]= 0.038435

And compare this by the CLT approximation
In[77]:= NProbability[Abs[z - 50 Mean[di]] >= Abs[221 - 50 Mean[di]], 
 z \[Distributed] 
  NormalDistribution[50 Mean[di], Sqrt[50 Variance[di]]]]

Out[77]= 0.0326175

A: Imagine recording the number of car troubles for $50$ days. Let random variable $X_1$ be the number on the first day, $X_2$ the number on the second, and so on up to $X_{50}$. Let $Y=X_1+\cdots+X_{50}$. Then $Y$ is the total number of car troubles in the $50$ day period.
We assume that the $X_i$ are independent and identically distributed, or at least not too far from that. These are implausible assumptions, but let's go on. Then $Y$ has approximately normal distribution, and $E(Y)=50E(X_1)$, $\operatorname{Var}(Y)=50\operatorname{Var}(X_1)$.  
The result about $E(Y)$ comes from the fact that by the linearity of expectation, we have $E(Y)=E(X_1+\cdots+X_{50})=E(X_1)+\cdots+E(X_{50})$
The result about $\operatorname{Var}(Y)$ comes from the fact that the variance of a sum of independent random variables is the sum of the variances. 
If your numerical calculations are correct (I have not verified them) then $Y$ has mean $206$, and variance $\approx 49.3$, so standard deviation $\approx 7$. 
The number $221$ is $15$ up from $206$, so a little more than $2$ standard deviation units. You can look up the probability that $Z\ge \frac{15}{7}$, where $Z$ is standard normal, and then reach a conclusion. 
Remark: The fact that the distribution of $Y$ is roughly normal tends to be justified by invoking the Central Limit Theorem. This is not really right. The CLT is a limit theorem. If $Y_n=X_1+\cdots+X_n$, it gives us information about the behaviour of $Y_n$ as $n\to\infty$. For any concrete situation, and the decidedly not very large $n=50$, the limiting behaviour is not directly relevant: we only care about thi particular distribution, and with $n=50$.  Knowing when the normal approximation is likely to be "good enough" is largely a matter of experience, and, sometimes, crossing one's fingers.  
