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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 vita\(\min \mathrm{E} ?\) at least \(5.2 \mathrm{mg} ?\)

Short Answer

Expert verified
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

01

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, \( Z = (5.2 - 5) / 0.05 = 4 \).
02

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%.
03

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.

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Key Concepts

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

Probability
Probability is an essential concept in statistics, representing the likelihood of an event happening. It is a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. When working with a normal distribution, calculating probabilities involves determining the likelihood of a random variable falling within a certain range.

For example, in the context of vitamin E capsules, we are interested in finding the probability that a capsule contains less than 4.9 mg or at least 5.2 mg. Using a standard normal distribution table or a calculator, we can find these probabilities based on our Z-scores.

Some key points about probability include:
  • It helps in making predictions based on data.
  • It uses statistical tools like standard normal distribution for calculations.
  • It's fundamental for making informed decisions in uncertain situations.
Standardization
Standardization is a process that allows us to compare different variables by converting them to a common scale. This involves transforming a variable into a Z-score using the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
Where \( X \) is the original value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. By doing this, we effectively rescale the variable so that the mean becomes 0 and the standard deviation becomes 1.

In the case of vitamin E capsules, standardization allows us to determine how far a capsule's content deviates from the average in standard deviation units. This is crucial as it simplifies probability calculations by referencing the standard normal distribution.

Reasons to standardize include:
  • Making different datasets comparable.
  • Facilitating the use of statistical tables.
  • Simplifying complex calculations.
Z-Score
Understanding Z-scores is key to working with normal distributions. A Z-score indicates how many standard deviations a data point is from the mean of its distribution.

In the formula
  • \( Z = \frac{X - \mu}{\sigma} \)
"\( X \)" is the value in question, "\( \mu \)" is the mean, and "\( \sigma \)" is the standard deviation.

When a z-score is calculated, it provides valuable insight into the positioning of \( X \) compared to the rest of the dataset. For instance, a Z-score of -2, which might indicate a vitamin E content of 4.9 mg, suggests the value is 2 standard deviations below the mean, implying it's quite below average.

Key takeaways about Z-scores are:
  • They help identify outliers.
  • They are critical for screening variability in data.
  • They allow for the use of standard normal distribution tables effectively.
Mean and Standard Deviation
The mean and standard deviation are paramount in understanding normal distribution. The mean is the average of all data points, serving as a central tendency measure. It represents where the data centers in the dataset. In the context of vitamin E capsules, it was 5 mg.

The standard deviation, on the other hand, measures the spread or dispersion of the data. It tells us how much the values deviate from the mean. In this example, the standard deviation was 0.05 mg, indicating minimal variation.

Knowing the mean and standard deviation allows us to predict the dataset's behavior and probability characteristics:
  • The mean provides context to deviations and variability.
  • The standard deviation quantifies how dispersed the values are.
  • Together, they define the shape and distribution features.

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Most popular questions from this chapter

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