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To determine the expected number of patients experiencing insomnia from the use of Clarinex-D, we first need to understand the concept of expected value. In probability and statistics, the expected value is a measure of the center of a probability distribution and is often referred to as the mean.
In this exercise, the probability of a patient experiencing insomnia is given as 0.05, or 5%. This implies that for each patient, there is a 5% chance that they will experience insomnia as a side effect. When dealing with a binomial distribution, the expected value can be calculated using the formula:
detailed as follows:
\[ E = n \times p \] where:

• \(E\) is the expected number of patients with side effects,


• \(n\) is the total number of trials (or patients, in this case),


• \(p\) is the probability of the event happening.

In this case:\[ E = 240 \times 0.05 = 12 \]So, we expect that 12 out of the 240 patients will experience insomnia as a side effect. This gives us an idea of the likely outcome based on the given probability.

Standard deviation is a measure of the amount of variation or dispersion in a set of values. For a binomial distribution, we use the formula:
\[ \text{Standard Deviation} = \sqrt{n \times p \times (1 - p)} \]where:

• \(n\) is the total number of trials (patients),


• \(p\) is the probability of the event happening (insomnia),


• \(1 - p\) is the probability of the event not happening.

In the context of our exercise, applying the given values:\[ n = 240 \] \[ p = 0.05 \]\[ \text{Standard Deviation} = \sqrt{240 \times 0.05 \times 0.95} \approx \sqrt{11.4} \approx 3.37 \]This means that the number of patients experiencing insomnia is expected to vary by about 3.37 from the expected value of 12. Standard deviation helps us understand how much the values can differ from the average.

The z-score is a statistical measure that tells us how many standard deviations an element is from the mean. To determine if it is unusual to observe 20 patients experiencing insomnia, we calculate the z-score using the formula:
\[ z = \frac{x - \mu }{\sigma} \]where:

• \(x\) is a specific value from the data set (number of patients with insomnia in this case),


• \(\mu\) is the mean (expected value),


• \(\sigma\) is the standard deviation.

For this problem:\[ x = 20 \]\[ \mu = 12 \]\[ \sigma = 3.37 \]By substituting these values into the formula:\[ z = \frac{20 - 12}{3.37} \approx \frac{8}{3.37} \approx 2.37 \]A z-score of approximately 2.37 tells us that 20 patients experiencing insomnia is 2.37 standard deviations above the mean. Since a z-score above 2 or below -2 is generally considered unusual, we conclude that 20 patients experiencing insomnia is statistically significant and less likely to occur based on the given probability.

Probability is a way of quantifying the likelihood of an event occurring. In this exercise, the probability of a patient experiencing insomnia while using Clarinex-D is given as 0.05 or 5%. This means that out of 100 patients, 5 are expected to experience insomnia.
In the context of binomial distribution, where there are only two possible outcomes (insomnia or no insomnia), we can determine the probability of any specific event, such as observing a certain number of patients with insomnia.
The concept of probability helps us understand not only the expected outcomes but also the variation and the chance of unusual outcomes. For example, while it is theoretically possible for all 240 patients to experience insomnia, it is extremely unlikely given the low probability of 5% per patient. This perspective allows us to make informed conclusions about the results of probability experiments.
In summary:

• Expected value gives a central tendency of the outcomes.


• Standard deviation tells us about the spread of the outcomes.


• Z-score helps us understand the unusualness of an outcome based on standard deviation.


• Probability grounds these results in real-world likelihood, helping us interpret them accurately.