To solve this problem, we can treat it as a binomial distribution scenario, where:

• The probability of success (a girl being born) is $p = 0.5$.
• The total number of trials (births in this case) is $n = 24$.
• The probability of failure (a boy being born) is $q = 1 - p = 0.5$.

a) Find the mean and the standard deviation for the number of girls in groups of 24 births.

1. Mean $\mu$ of a binomial distribution is given by: $\mu = n \times p$ Substituting the given values: $\mu = 24 \times 0.5 = 12$ So, the mean is 12.

2. Standard deviation $\sigma$ is calculated as: $\sigma = \sqrt{n \times p \times q}$ Substituting the values: $\sigma = \sqrt{24 \times 0.5 \times 0.5} = \sqrt{6} \approx 2.449$ So, the standard deviation is approximately 2.449.

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Related Questions:

1. What is the probability of getting exactly 12 girls in 24 births?
2. How would the standard deviation change if the probability of a girl being born was 0.6?
3. What is the probability of getting more than 16 girls in 24 births?
4. How would the mean change if there were 30 births instead of 24?
5. How would you interpret the standard deviation in the context of this problem?

Tip:

When dealing with binomial distributions, remember that the standard deviation shows how much the actual results are likely to vary from the mean in repeated experiments.