Question

the probability of flu symptoms for a person not receiving any treatment is 0.058. in a clinical trial of a common drug used to lower cholesterol, 66 of 1092 people treated experienced flu symptoms. assuming the drug has no effect on the likelihood of flu symptoms, estimate the probability that at least 66 people experience flu symptoms. what do these results suggest about flu symptoms as an adverse reaction to the drug?
(a) p(x ≥ 66) = (round to four decimal places as needed.)

Explanation:

Step1: Identify distribution parameters

This is a binomial - to - normal approximation problem. Let \(n = 1092\) (number of people treated) and \(p=0.058\). The mean of the binomial distribution is \(\mu = np\) and the standard deviation is \(\sigma=\sqrt{np(1 - p)}\).
\(\mu=np=1092\times0.058 = 1092\times0.058=63.336\)
\(\sigma=\sqrt{np(1 - p)}=\sqrt{1092\times0.058\times(1 - 0.058)}=\sqrt{63.336\times0.942}=\sqrt{59.662512}\approx7.7241\)

Step2: Apply continuity correction

For \(P(X\geq66)\) in a binomial distribution approximated by a normal distribution, we use the continuity correction. So we find \(P(X\geq65.5)\) for the normal distribution.

Step3: Standardize the value

We use the z - score formula \(z=\frac{x-\mu}{\sigma}\), where \(x = 65.5\), \(\mu = 63.336\) and \(\sigma\approx7.7241\).
\(z=\frac{65.5 - 63.336}{7.7241}=\frac{2.164}{7.7241}\approx0.28\)

Step4: Calculate the probability

\(P(X\geq65.5)=1 - P(X\lt65.5)=1 - P(Z\lt0.28)\)
From the standard normal table, \(P(Z\lt0.28) = 0.6103\)
\(P(X\geq65.5)=1 - 0.6103=0.3897\)

Answer:

\(0.3897\)

These results suggest that if the drug has no effect on the likelihood of flu - symptoms, the probability of having at least 66 people with flu - symptoms is relatively high (\(0.3897\)). This indicates that the observed number of people with flu - symptoms may not be strong evidence that flu - symptoms are an adverse reaction to the drug. It could be due to normal random variation in a population with a baseline flu - symptom probability of \(0.058\).