To solve these problems, we can use the binomial probability formula, as we are dealing with a fixed number of independent trials (20 people) and a constant probability of success (11.5% or 0.115). For each part:
a) Use the binomial probability formula to find the probability of exactly 4 successes.
b) Sum the probabilities of having 0 or 1 success to find the probability of fewer than 2 successes.
c) Calculate the probability of having more than 3 successes by subtracting the probability of having 3 or fewer successes from 1.
To find the probability that exactly 4 out of 20 people suffer from chronic sinusitis, we use the binomial probability formula:
\[
P(X = 4) = \binom{20}{4} \cdot (0.115)^4 \cdot (1 - 0.115)^{16}
\]
The calculated probability is approximately \(0.1200\).
To find the probability that fewer than 2 people suffer from chronic sinusitis, we sum the probabilities of having 0 or 1 success:
\[
P(X < 2) = P(X = 0) + P(X = 1)
\]
The cumulative probability is approximately \(0.3126\).
To find the probability that more than 3 people suffer from chronic sinusitis, we subtract the probability of having 3 or fewer successes from 1:
\[
P(X > 3) = 1 - P(X \leq 3)
\]
The calculated probability is approximately \(0.1914\).
- The probability that exactly 4 in the sample suffer from chronic sinusitis is \(\boxed{0.1200}\).
- The probability that there are fewer than 2 in the sample that suffer from chronic sinusitis is \(\boxed{0.3126}\).
- The probability that there are more than 3 in the sample that suffer from chronic sinusitis is \(\boxed{0.1914}\).