CLT: Cheese

Joseph Harding loves cheese and decides to make 100 blocks of cheese. The weight of each block follows an independently and identically distributed exponential distribution with a mean of 250 grams. Let $$W$$ represent the weight of the $$i$$ -th block of cheese, and $T_{100}$ be the total weight of the 100 blocks of cheese. Using the Central Limit Theorem, can you approximate the probability $P(T_{100} > 26000)$ ? This probability can be expressed in the form $\Phi(\alpha)$ for some real number $\alpha$ . What is the value of $\alpha$ ?

Answer

To find the approximation for $P(T_{100} > 26000)$ , we start by subtracting the expected total from 26000 (since the expected total of $T_{100}$ is $100 \times 250 = 25000$ ). Then we divide by the standard deviation of $T_{100}$ , calculated as $\sqrt{100 \times (250)^2} = 2500$ . This gives us:

P(T_{100} > 26000) \;=\; 1 \;-\; P\bigl(T_{100} < 26000\bigr) \;=\; 1 \;-\; P\!\Bigl(Z < \frac{26000 - 25000}{2500}\Bigr) \;=\; 1 \;-\; \Phi\!\Bigl(\frac{1000}{2500}\Bigr).

We approximate the left-hand side using a standard normal random variable by the Central Limit Theorem. So we are essentially finding $1 - \Phi\bigl(\tfrac{2}{5}\bigr)$ , which is equivalent to $\Phi\bigl(-\tfrac{2}{5}\bigr)$ . Therefore, $\alpha = -\tfrac{2}{5}$ .

Notes and comments

Comment 1: The Central Limit Theorem (CLT) asserts that if $X_1, X_2, \ldots, X_n$ are i.i.d. random variables with mean $\mu$ and standard deviation $\sigma$ , then

\frac{S_n - n\mu}{\sqrt{n}\sigma} \quad \text{converges in distribution to } N(0,1).

Equivalently, since the sample mean is given by $\overline{X}_n = \frac{S_n}{n}$ , we have

\frac{\overline{X}_n - \mu}{\sigma/\sqrt{n}} \quad \text{converges in distribution to } N(0,1).