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

Short Answer

Expert verified
There is a 2.28% chance that a randomly selected capsule contains less than 4.9 mg of vitamin E and a 0.01% chance that it contains at least 5.2 mg of vitamin E.

Step by step solution

01

Define the variables

The actual amount of vitamin E is normally distributed with a mean (µ) of 5 mg and a standard deviation (σ) of 0.05 mg. We are asked to find the probability of a randomly selected capsule containing less than 4.9 mg and at least 5.2 mg of vitamin E.
02

Calculate the z-scores

First, we'll find the z-score for 4.9 mg: Z-score = \(\frac{X - \mu}{\sigma}\) Where X is the value we want to find the probability for (in this case 4.9 mg), µ is the mean, and σ is the standard deviation. For 4.9 mg: \(Z_1 = \frac{4.9 - 5}{0.05}\) \(Z_1 = -2\) Second, we'll find the z-score for 5.2 mg: For 5.2 mg: \(Z_2 = \frac{5.2 - 5}{0.05}\) \(Z_2 = 4\)
03

Use the standard normal table

Now, we'll check the standard normal table (z-table) to find the probabilities associated with these z-scores. For Z = -2, the probability is 0.0228, which means there is a 2.28% chance that a randomly selected capsule contains less than 4.9 mg of vitamin E. For Z = 4, the probability is 0.9999 or 99.99%. However, we need the probability of having at least 5.2 mg, which means we'll take the complement (1 minus the probability): 1 - 0.9999 = 0.0001 or 0.01%
04

State the results

There is a 2.28% chance that a randomly selected capsule contains less than 4.9 mg of vitamin E and a 0.01% chance that it contains at least 5.2 mg of vitamin E.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Distribution
Imagine you're at a party and a bowl is filled with vitamin E capsules. Not all capsules contain the same amount of vitamin E, and you're curious about the chances of grabbing one with a certain amount. A probability distribution is like a list that shows you how likely it is to pick capsules with different amounts of vitamin E.

For example, most capsules might have around 5 mg of vitamin E, which is the 'average' amount you expect to find. But the reality is, some capsules will have a bit more or less. A probability distribution would show you the odds or 'probability' for each possible amount, from the least likely to the most likely. The normal distribution, which applies to our vitamin E capsules, describes a scenario where the most common outcomes cluster around the mean (in our case, 5 mg), and outcomes become less common the further they are from the mean. This is represented graphically by a bell curve, with the peak at the mean. The higher the curve at any given point, the higher the likelihood of a capsule having that amount of vitamin E.
Z-Score Calculation
Now, suppose you want to know how 'unusual' it is to find a capsule with 4.9 mg or 5.2 mg of vitamin E. A z-score can help with this! It's a way of measuring how far and in what direction, a value (like 4.9 mg) is from the mean, considering the spread of the values (which is the standard deviation).

To calculate the z-score, you take the value you're investigating (let's say, 4.9 mg), subtract the average amount (5 mg), and then divide by the standard deviation (0.05 mg). So, when you do that for 4.9 mg, you get \( Z_1 = \frac{4.9 - 5}{0.05} \), which equals -2. This means 4.9 mg is two standard deviations below the mean. Negative here means it's less than the average amount of vitamin E. Likewise, for a capsule with 5.2 mg, the z-score is +4, showing it's far above average. So, z-scores simply tell us how typical or atypical certain vitamin E amounts are, using the common language of 'standard deviations from the mean'.
Standard Normal Table
Armed with z-scores, we still need to know what the chances are of picking those special capsules. This is where the standard normal table, or z-table, comes into play. It's like a treasure map that helps us find the treasure of probability values.

How does it work? Well, you take your z-score and look it up in the table. For our vitamin E example, a z-score of -2 corresponds to a probability value in the table. This tells us that if you pick a capsule randomly, there's about 2.28% chance it'll have less than 4.9 mg of vitamin E. On the other hand, a z-score of +4 is so high that it's not typically listed, but we can assume it's close to 100%, which means almost all capsules will have less than 5.2 mg of vitamin E. To find the probability of at least 5.2 mg, we subtract this value from 1, giving us a tiny 0.01% chance. So, the standard normal table translates z-scores into probabilities, giving us a powerful tool for understanding and predicting the outcomes.

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