To solve this problem, we can model it using the binomial probability distribution. Here's the approach:

• n is the number of trials (wolves spotted), which is 11.
• p is the probability of success (male wolves), which is 0.55 before 1918.
• We are asked to find the probability that 8 or more wolves are male in a sample of 11 wolves.

The binomial probability formula is:

$P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k}$

Where:

• $X$ is the number of successes (male wolves).
• $\binom{n}{k}$ is the binomial coefficient, which represents the number of ways to choose $k$ successes in $n$ trials.
• $p^k$ is the probability of $k$ successes.
• $(1 - p)^{n - k}$ is the probability of $n - k$ failures.

We need to calculate the probability of getting 8 or more male wolves, so we sum the probabilities for $X = 8, 9, 10,$ and $11$:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11)$

Let’s compute this using the binomial distribution.

The probability that 8 or more wolves in a random sample of 11 were male before 1918 is approximately 0.191 (or 19.1%).

Do you want further details on the steps or any clarifications? Here are some related questions to think about:

1. What is the probability that exactly 7 wolves were male before 1918?
2. How does the probability change if the sample size increases to 15 wolves?
3. What is the probability of spotting 10 or more male wolves in a sample of 11 after 1918?
4. How does the binomial distribution compare to a normal approximation for large sample sizes?
5. Can we calculate the expected number of male wolves in a sample of 20 before 1918?

Tip: The binomial distribution is a discrete distribution used when there are only two possible outcomes (success or failure) in a fixed number of trials.