Worked example

Linear Combination and CLT Worked Example

This example combines two of the biggest Exam P habits: get the mean and variance of a sum right, then standardize cleanly when the CLT gives you a normal approximation.

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clt worked example exam p

Problem

Suppose claim counts X1, ..., X100 are iid with mean 2 and variance 3. Use the Central Limit Theorem to approximate P(X1 + ... + X100 > 230).

What This Example Is Testing

This is testing whether you can move from a sum of iid random variables to its approximate normal distribution. The CLT only becomes useful if you first compute the mean and variance of the sum correctly.

Step-By-Step Solution

Let S100=X1+...+X100. Because the variables are iid, E[S100]=100 x 2=200 and Var(S100)=100 x 3=300. So the standard deviation is sqrt(300), about 17.32.

By the Central Limit Theorem, S100 is approximately normal with mean 200 and variance 300. Standardize the threshold 230: z=(230-200)/sqrt(300)=30/17.32, about 1.73.

Now use the standard normal tail. P(S100 > 230) is approximately P(Z > 1.73), which is about 0.0418.

Mean and variance of the sum
E[Sn]=nμ,Var(Sn)=nσ2E[S_n]=n\mu, \qquad \operatorname{Var}(S_n)=n\sigma^2
CLT standardization
Z=SnnμσnZ=\frac{S_n-n\mu}{\sigma\sqrt{n}}

Final Answer

The CLT approximation is P(X1 + ... + X100 > 230) \approx 0.0418.

Common Wrong Answer

A common wrong answer is using the wrong variance for the sum, usually by forgetting that variances add for independent variables. Another is plugging 300 in as the standard deviation instead of taking the square root first.

References And Official Sources