Question

A machine producing vitamin E capsules operates so that the actual amount of vitamin \(\mathrm{E}\) in each capsule is normally distributed with a mean of \(5 \mathrm{mg}\) and a standard deviation of \(0.05 \mathrm{mg} .\) What is the probability that a randomly selected capsule contains less than \(4.9 \mathrm{mg}\) of vita\(\min \mathrm{E} ?\) at least \(5.2 \mathrm{mg} ?\)

Short answer

The probability that a randomly selected capsule contains less than 4.9mg of vitamin E is approximately 2.28%. The probability that it contains at least 5.2mg of vitamin E is approximately 0.003%.

Step-by-step solution

Standardization

To start, one must first standardize the given quantities (4.9mg and 5.2mg), transforming them into z-scores, using the standardization formula \( Z = (X - \mu) / \sigma \), where \( X \) is the given quantity to be standardized, \( \mu \) is the mean and \( \sigma \) is the standard deviation. For 4.9mg, \( Z = (4.9 - 5) / 0.05 = -2 \). For 5.2mg, \( Z = (5.2 - 5) / 0.05 = 4 \).

Determine the Probability

Then, one should look up the z-scores in a standard normal distribution table or use a statistical software or calculator to find the corresponding probabilities. For \( Z = -2 \), the probability \( P(Z < -2) \) is approximately 0.0228, which means there is a 2.28% chance that a capsule has less than 4.9mg of vitamin E. For \( Z = 4 \), the probability \( P(Z < 4) \) is approximately 0.99997 or nearly 100%.

Calculating relevant probabilities

Finally, as the question asks for the probability that a capsule will have at least 5.2mg, one needs to subtract the probability associated with 5.2mg from 1 (as it's a case of 'more than'), to find the relevant probability. So, \( P(Z \geq 4) = 1 - P(Z < 4) = 1 - 0.99997 = 0.00003 \). This indicates there is a 0.003% chance that a capsule has at least 5.2mg of vitamin E.