Question

A machine producing vitamin \(\mathrm{E}\) 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 vitamin \(\mathrm{E}\) ? At least \(5.2 \mathrm{mg}\) of vitamin \(\mathrm{E}\) ?

Short answer

The probability that a randomly selected capsule contains less than 4.9 mg of vitamin E is 2.28%. The probability that a capsule contains at least 5.2 mg of vitamin E is effectively 0%.

Step-by-step solution

Understand and apply Z-Score formula

Z-Score is the value which shows how many standard deviations an element is from the mean. It is calculated using the formula \(Z = (X - μ) / σ\), where X is the value, μ is the mean, and σ is the standard deviation.

Calculate Z-Score for 4.9 mg

For the case of 4.9 mg of Vitamin E, substitute X = 4.9, μ = 5, and σ = 0.05 into the formula to get \(Z_1 = (4.9 - 5) / 0.05 = -2\).

Use Standard Normal Distribution to find the associated probability

The corresponding probability for \(Z = -2\) from the standard normal distribution table (or a calculator with this function) is 0.0228. That means the probability that a randomly selected capsule contains less than 4.9 mg of Vitamin E is 0.0228 or 2.28%.

Calculate Z-Score for 5.2 mg

By the same method for the case of 5.2 mg of Vitamin E, substitute X = 5.2, μ = 5, and σ = 0.05 into the formula to get \(Z_2 = (5.2 - 5) / 0.05 = 4\).

Use Standard Normal Distribution to find the associated probability

By getting the corresponding value for \(Z = 4\) from the standard normal distribution table (or a calculator with this function), it is seen that this represents almost the entire area under the curve (effectively 100%), so the probability of less than 5.2 mg is close to 1. Thus, the probability of at least 5.2 mg is 1 - this probability, which is close to 0. So it's extremely unlikely to get a capsule with 5.2 mg or higher of Vitamin E.

Key concepts

Z-Score Calculation

The Z-Score is an essential concept in statistics, representing the number of standard deviations a data point is from the mean of the distribution. It transforms data from a normal distribution into the standard normal distribution, allowing for easy comparison of different data points.

The formula for calculating the Z-Score is: \( Z = (X - \mu) / \sigma \) where \( X \) is the value in question, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. Getting familiar with this calculation can significantly simplify understanding how unusual a particular observation is within the context of the data set.

For example, if we have vitamin E capsules with a mean content of \( 5 mg \) and a standard deviation of \( 0.05 mg \) and we want to determine the Z-Score for a capsule with \( 4.9 mg \) of vitamin E, we simply substitute these values into our formula. With \( X = 4.9 mg \) the Z-Score comes out to be \( -2 \) indicating that this capsule has \( 2 \) standard deviations less vitamin E than the average capsule.

It’s also worth noting that positive Z-Score values indicate a measurement above the mean, while negative values are below the mean.

Standard Normal Distribution

The standard normal distribution is a special case of the normal distribution that is centered at zero and has a standard deviation of one. This distribution is crucial because it allows us to convert any normal distribution to a common scale using the Z-Score, enabling us to use standard normal distribution tables or relevant software to find probabilities and percentiles.

Since the standard normal distribution is symmetrical, the table or calculator provides the probability for values to the left of the Z-Score. Upon finding the Z-Score from our previous Z-Score calculation, we can determine the likelihood of a data point falling below (or above) that Z-Score. For instance, a Z-Score of \( -2 \) has a corresponding probability of \( 0.0228 \) in the standard normal distribution, indicating a \( 2.28\% \) chance that a randomly selected capsule contains less than \( 4.9 mg \) of vitamin E.

Standard Deviation

Standard deviation is a measure of spread or dispersion within a distribution. In simpler terms, it quantifies how much individual data points differ from the mean value. For a normal distribution, nearly all data will fall within three standard deviations of the mean.

Understanding standard deviation is pivotal for interpreting Z-Scores since it forms the denominator in the calculation. A smaller standard deviation implies that the data points are tightly clustered around the mean, while a larger standard deviation indicates more spread. For the vitamin E capsules, a standard deviation of \( 0.05 mg \) suggests that the content of most capsules is quite consistent, rarely deviating from the \( 5 mg \) mean by more than a small margin.

Therefore, when you encounter a high or low Z-Score in your calculations, analyze the standard deviation to understand the context. If the standard deviation is small and the Z-Score is large, the data point is exceptionally unusual; conversely, a large standard deviation could mean that such deviations are common and perhaps less remarkable.