Answered

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 patients in the study experienced insomnia as a side effect of taking the medication. Suppose a random sample of 20 patients who took Clarinex-D is obtained and the number of patients who experienced insomnia, X, is recorded. For parts (b)-(d), write a probability statement, such as P(X>3), and calculate the probability (either by hand or using software). Be careful of the wording. Know the difference between less than, at most, etc.

Required:
a. What are the possible values of X?
b. Find the probability that exactly 3 experience insomnia as a side effect?
c. Find the probability 3 or fewer experience insomnia?
d. Between 1 and 4 experience insomnia.
e. Would it be unusual to find 4 or more patients experience insomnia as a side effect?



Answer :

Using the binomial distribution, it is found that:

a) The possible values of X are between 0 and 20.

b) 0.0596 = 5.96% probability that exactly 3 experience insomnia as a side effect.

c) 0.9842 = 98.42% probability 3 or fewer experience insomnia.

d) 0.639 = 63.9% probability that between 1 and 4 experience insomnia.

e) Since [tex]4 > E(X) + 2.5\sqrt{V(X)}[/tex], it would be unusual to find 4 or more patients experience insomnia as a side effect.

For each patient, there are only two possible outcomes, either they experience insomnia, or they did not. The probability of a patient having experienced insomnia is independent of any other patient, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

Probability of exactly x successes on n trials, with p probability.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]  

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]  

The expected value is:

[tex]E(X) = np[/tex]

The standard deviation is:

[tex]\sqrt{V(X)} = \sqrt{np(1 - p)}[/tex]

If a measure is more than 2.5 standard deviations above the mean, it is considered unusual.

In this problem:

  • 20 trials, thus [tex]n = 20[/tex]
  • 5% experience insomnia, thus [tex]p = 0.05[/tex]

Item a:

The possible values of X are always between 0 and n, thus in this case, it is between 0 and 20.

Item b:

This probability is P(X = 3), thus:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 3) = C_{20,3}.(0.05)^{3}.(0.95)^{17} = 0.0596[/tex]

0.0596 = 5.96% probability that exactly 3 experience insomnia as a side effect.

Item c:

This probability is [tex]P(X \leq 3)[/tex], given by:

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

Then:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{20,0}.(0.05)^{0}.(0.95)^{20} = 0.3585[/tex]

[tex]P(X = 1) = C_{20,1}.(0.05)^{1}.(0.95)^{19} = 0.3774[/tex]

[tex]P(X = 2) = C_{20,2}.(0.05)^{2}.(0.95)^{18} = 0.1887[/tex]

[tex]P(X = 3) = C_{20,3}.(0.05)^{3}.(0.95)^{17} = 0.0596[/tex]

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3585 + 0.3774 + 0.1887 + 0.0596 = 0.9842[/tex]

0.9842 = 98.42% probability 3 or fewer experience insomnia.

Item d:

This probability is:

[tex]P(1 \leq X \leq 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]

Then

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 1) = C_{20,1}.(0.05)^{1}.(0.95)^{19} = 0.3774[/tex]

[tex]P(X = 2) = C_{20,2}.(0.05)^{2}.(0.95)^{18} = 0.1887[/tex]

[tex]P(X = 3) = C_{20,3}.(0.05)^{3}.(0.95)^{17} = 0.0596[/tex]

[tex]P(X = 4) = C_{20,4}.(0.05)^{4}.(0.95)^{16} = 0.0133[/tex]

[tex]P(1 \leq X \leq 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.3774 + 0.1887 + 0.0596 + 0.0133 = 0.639[/tex]

0.639 = 63.9% probability that between 1 and 4 experience insomnia.

Item e:

[tex]E(X) = np = 20(0.05) = 1[/tex]

[tex]\sqrt{V(X)} = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.95)} = 0.975[/tex]

[tex]E(X) + 2.5\sqrt{V(X)} = 1 + 2.5(0.975) = 3.44[/tex]

Since [tex]4 > E(X) + 2.5\sqrt{V(X)}[/tex], it would be unusual to find 4 or more patients experience insomnia as a side effect.

A similar problem is given at https://brainly.com/question/24863377