Question

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?

Answer 1

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:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

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

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

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 1164, p = 0.038$

So

$\mu = E(X) = np = 1164*0.038 = 44.232$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = 6.5231$

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

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{46.5 - 44.232}{6.5231}$

$Z = 0.35$

$Z = 0.35$ 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.