The probability of flu symptoms for a person not receiving any treatment is 0.038. In a clinical trial of a common drug used to lower​ cholesterol, 47 of 1164 people treated experienced flu symptoms. Assuming the drug has no effect on the likelihood of flu​ symptoms, estimate the probability that at least 47 people experience flu symptoms. What do these results suggest about flu symptoms as an adverse reaction to the​ drug?

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Answer:

36.32% probability that at least 47 people experience flu symptoms. This is not an unlikely event, so this suggests that flu symptoms are not an adverse reaction to the drug.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

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

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

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

The standard deviation of the binomial distribution is:

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

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 1164, p = 0.038[/tex]

So

[tex]\mu = E(X) = np = 1164*0.038 = 44.232[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = 6.5231[/tex]

Estimate the probability that at least 47 people experience flu symptoms.

Using continuity correction, this is [tex]P(X \geq 47 - 0.5) = P(X \geq 46.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 46.5.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{46.5 - 44.232}{6.5231}[/tex]

[tex]Z = 0.35[/tex]

[tex]Z = 0.35[/tex] has a 0.6368

1 - 0.6368 = 0.3632

36.32% probability that at least 47 people experience flu symptoms. This is not an unlikely event, so this suggests that flu symptoms are not an adverse reaction to the drug.