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Chapter 6: Problem 46

Clarinex-D is a medication whose purpose is to reduce the symptoms associated with a variety of allergies. In clinical trials of Clarinex-D, \(5 \%\) of the patients in the study experienced insomnia as a side effect. (a) If 240 users of Clarinex-D are randomly selected, how many would we expect to experience insomnia as a side effect? (b) Would it be unusual to observe 20 patients experiencing insomnia as a side effect in 240 trials of the probability experiment? Why?

Short Answer

Expert verified
We expect 12 out of 240 patients to experience insomnia. Observing 20 patients with insomnia is unusual because it exceeds the typical range of 5.24 to 18.76 patients.

Step by step solution

01

- Identify given values

The problem states that 5% of the patients experienced insomnia. This percentage can be converted to a decimal for calculation purposes: \( p = 0.05 \). The total number of patients is given as 240, so: \( n = 240 \).
02

- Calculate the expected number of patients with insomnia

To find the expected number of patients experiencing insomnia, multiply the total number of patients by the probability: \( E = n \times p \). Substituting the given values: \( E = 240 \times 0.05 = 12 \). Thus, we would expect 12 patients to experience insomnia.
03

- Determine if observing 20 patients with insomnia is unusual

To determine if observing 20 patients with insomnia is unusual, we use the binomial distribution's properties. First, calculate the mean (\mu) and standard deviation (\sigma) of the distribution: \( \mu = n \times p = 240 \times 0.05 = 12 \) \( \sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{240 \times 0.05 \times 0.95} = \sqrt{11.4} \approx 3.38 \). An outcome is considered unusual if it lies more than 2 standard deviations from the mean. Calculate the range of usual values: \( 12 \pm 2 \times 3.38 = 12 \pm 6.76 \). So, the usual range is approximately from 5.24 to 18.76 patients. Since 20 patients lie outside of this range, it would be unusual to observe 20 patients experiencing insomnia.

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Key Concepts

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

Expected Value Calculation
In the context of the provided problem, the expected value tells us the average number of Clarinex-D users (out of 240) we should expect to experience insomnia due to the medication. To find the expected value (E), we use the formula:
\(E = n \times p\).
Here, \(n\) represents the number of trials or users, which is 240. And \(p\) represents the probability of experiencing insomnia, which is 0.05.
By substituting these values, we get:
\(E = 240 \times 0.05 = 12\).
This means we expect, on average, 12 out of 240 users to experience insomnia.
Standard Deviation in Binomial Distribution
Standard deviation is crucial because it provides insight into the variability or spread of the outcomes around the mean. For a binomial distribution, the standard deviation (蟽) is calculated using the formula:
\(\frac{\text{{蟽}}}{\text{{Standard Deviation}}} = \text{{鈭歔n \times p \times (1 - p)]}}\).
Substituting the given values, where \(n = 240\) and \(p = 0.05\), we obtain:
\(\text{{蟽}} = \text{{鈭歔240 \times 0.05 \times 0.95]}} = \text{{鈭歔11.4]}} \text{{鈮� 3.38}}\).
Standard deviation helps determine the range in which the number of patients experiencing insomnia will most likely fall.
Probability of Unusual Events
To determine if 20 patients out of 240 experiencing insomnia is unusual, we analyze how far this value is from the expected mean (E) using the standard deviation (蟽). An outcome is considered unusual if it lies more than 2 standard deviations from the mean.
First, calculate the range of 'usual' values:
\( \text{{Mean}} \text{{卤 2蟽}} = \text{{12 卤 (2 脳 3.38)}} = \text{{12 卤 6.76}}\).
The usual range therefore spans from approximately 5.24 to 18.76 patients.
Since 20 patients fall outside this range, it would be considered unusual. This calculation helps us understand the likelihood and rarity of certain outcomes within a binomial probability experiment.

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Most popular questions from this chapter

Connecticut Lottery In the Cash Five Lottery in Connecticut, a player pays \(\$ 1\) for a single ticket with five numbers. Five balls numbered 1 through 35 are randomly chosen from a bin without replacement. If all five numbers on a player's ticket match the five chosen, the player wins \(\$ 100,000 .\) The probability of this occurring is \(\frac{1}{324,632} .\) If four numbers match, the player wins \(\$ 300 .\) This occurs with probability \(\frac{1}{2164} .\) If three numbers match, the player wins \(\$ 10 .\) This occurs with probability \(\frac{1}{75}\). Compute and interpret the expected value of the game from the player's point of view.

According to www.meretrix.com, airline fatalities occur at the rate of 0.05 fatal accidents per 100 million miles. Find the probability that, during the next 100 million miles of flight, there will be (a) exactly zero fatal accidents. Interpret the result. (b) at least one fatal accident. Interpret the result. (c) more than one fatal accident. Interpret the result.

A probability distribution for the random variable \(X,\) the number of trials until a success is observed, is called the geometric probability distribution. It has the same criteria as the binomial distribution (see page 328 ), except that the number of trials is not fixed. Its probability distribution function (pdf) is $$ P(x)=p(1-p)^{x-1}, \quad x=1,2,3, \ldots $$ where \(p\) is the probability of success. (a) What is the probability that Shaquille O'Neal misses his first two free throws and makes the third? Over his career, he made \(52.4 \%\) of his free throws. That is, find \(P(3)\) (b) Construct a probability distribution for the random variable \(X,\) the number of free-throw attempts of Shaquille O'Neal until he makes a free throw. Construct the distribution for \(x=1,2,3, \ldots, 10 .\) The probabilities are small for \(x>10\) (c) Compute the mean of the distribution, using the formula presented in Section 6.1 (d) Compare the mean obtained in part (c) with the value \(\frac{1}{p}\). Conclude that the mean of a geometric probability distribution is \(\mu_{X}=\frac{1}{p} .\) How many free throws do we expect Shaq to take before we observe a made free throw?

According to flightstats.com, American Airlines flights from Dallas to Chicago are on time \(80 \%\) of the time. Suppose 15 flights are randomly selected, and the number of on-time flights is recorded. (a) Explain why this is a binomial experiment. (b) Find and interpret the probability that exactly 10 flights are on time. (c) Find and interpret the probability that fewer than 10 flights are on time. (d) Find and interpret the probability that at least 10 flights are on time. (e) Find and interpret the probability that between 8 and 10 flights, inclusive, are on time.

(a) construct a binomial probability distribution with the given parameters; (b) compute the mean and standard deviation of the random variable using the methods of Section \(6.1 ;\) (c) compute the mean and standard deviation, using the methods of this section; and \((d)\) draw a graph of the probability distribution and comment on its shape. $$ n=10, p=0.5 $$
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