Some tablets of a company are refilled by a machine

Question

Some tablets of a company are refilled by a machine that each contain 25 mg of medication. The amount that goes into the tablets is normally distributed, with mean 25 mg and a standard deviation of 5 mg.

What is the probability that the total weight of medication in 20 randomly selected tablets will be at least 500 mg?

I was wondering if such a question would require the Central Limit Theorem, or can it be done some other way?

Thank you.

Answer 1

The central limit theorem does not need to be invoked, because you are told that the medication amount in each capsule is normally distributed. Therefore, the total medication contained in ten capsules is also normally distributed, without the need for the CLT, because the sum of independent normal random variables is also normal.

Using your notation, $$X_i \sim \operatorname{Normal}(\mu = 10, \sigma = 0.5), \quad i = 1, 2, \ldots 10,$$ hence $$S_{10} = \sum_{i=1}^{10} X_i \sim \operatorname{Normal}(\mu_S = 100, \sigma_S = 0.5 \sqrt{10}).$$ It follows that $$\Pr[S_{10} \ge 90] = \Pr\left[\frac{S_{10} - \mu_S}{\sigma_S} \ge \frac{90 - 100}{0.5 \sqrt{10}} \right].$$ What we did in the step above was standardize $S_{10}$. The LHS of the inequality is now a standard normal random variable $Z$ with zero mean and unit variance; thus $$\Pr[S_{10} \ge 90] = \Pr[Z \ge -6.32456].$$ The probability that a standard normal random variable is greater than $-6.32456$ standard deviations below its mean of $0$ is almost equal to $1$; using a computer, it is $$0.999999999873018570526456\ldots.$$

Think about why this answer makes sense. If the standard deviation of a single capsule is $0.5$ mg, and you have $10$ capsules, in order for the total medication amount to be $10$ mg below the total mean, ($90$ is $10$ below the mean of $100$), each capsule would on average need to contain at least $2$ standard deviations below the average amount per capsule. Since being $2$ or more standard deviations below the mean happens only about $2.27\%$ of the time, it should not be surprising that the answer is almost $1$.