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The problem states:

- About 1% of the population has a particular genetic mutation.
- 300 people are randomly selected.
- We are tasked with finding the mean for the number of people with the genetic mutation in such a group of 300.

This situation can be modeled using a binomial distribution because each person either has the mutation or does not, and the trials (people selected) are independent.

In a binomial distribution, the mean \(\mu\) is calculated as:
\[
\mu = n \cdot p
\]
Where:
- \(n = 300\) (the number of people selected),
- \(p = 0.01\) (the probability that a person has the genetic mutation).

Let’s compute the mean.

\[
\mu = 300 \times 0.01 = 3
\]

Thus, the mean number of people with the genetic mutation in a group of 300 is 3.

Would you like more details or have any questions?

Here are 5 related questions you might consider:
1. How would the variance and standard deviation of the distribution be calculated?
2. What is the probability that no one in the group of 300 has the mutation?
3. How would you model this if the percentage of people with the mutation was different, say 5%?
4. Can this be modeled using a Poisson distribution? Under what conditions would this be appropriate?
5. How would the result change if 500 people were selected instead of 300?

**Tip**: For large sample sizes and small probabilities, the binomial distribution can be approximated by the Poisson distribution to simplify calculations.

About 1% of the population has a particular genetic mutation. 300 people are randomly selected. Find the mean for the number of people with the genetic mutation in such a group of 300.

Learn how to calculate the mean number of people with a genetic mutation in a group of 300, given that 1% of the population carries the mutation. This problem uses the binomial distribution to model the situation. Suitable for students in Grades 10-12.

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