Let's solve the problem step by step.

Problem Breakdown:

• 22% of the fish in the lake have been tagged.
• A sample of 12 fish is captured and released, with replacement.
• $X$ is the number of tagged fish captured.

Part a) Specify the distribution of $X$:

Since each fish is captured with replacement, and each fish has a fixed probability (22%) of being tagged, the number of tagged fish in the sample follows a binomial distribution.

• Parameters of the binomial distribution:
  • $n = 12$ (number of trials, i.e., number of fish captured)
  • $p = 0.22$ (probability of capturing a tagged fish)

Thus, the distribution of $X$ is: $X \sim \text{Binomial}(n = 12, p = 0.22)$

Part b) Expected value of $X$:

For a binomial distribution, the expected value $E(X)$ is given by: $E(X) = n \cdot p$ Substituting the values: $E(X) = 12 \cdot 0.22 = 2.64$

So, the expected value is 2.6 (rounded to one decimal place).

Part c) Probability of getting 4 or more tagged fish $P(X \geq 4)$:

To compute $P(X \geq 4)$, we can use the complement rule: $P(X \geq 4) = 1 - P(X \leq 3)$ We need to compute the cumulative probability for $X \leq 3$ using the binomial formula: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ For $X \leq 3$, we calculate the probabilities for $X = 0, 1, 2, 3$, and sum them. Then subtract the sum from 1 to find $P(X \geq 4)$.

I’ll calculate this now.The probability of capturing 4 or more tagged fish $P(X \geq 4)$ is approximately 0.261 (rounded to three decimal places).

Would you like more details or have any questions about this solution?

Related questions:

1. What is the variance of the binomial distribution in this case?
2. How would the results change if the number of captured fish increased to 20?
3. What is the probability of capturing exactly 2 tagged fish?
4. Can we approximate this binomial distribution with a normal distribution?
5. How would the probability change if 30% of the fish were tagged instead?

Tip:

When working with binomial distributions, always verify if the sample size $n$ and the probability $p$ allow for approximations using other distributions, like the normal distribution, especially when $n$ is large.