Since we are given that each couple has a probability of 0.5 of having a girl, this scenario can be modeled by a binomial distribution.

The parameters of a binomial distribution are:

• $n = 20$ (the number of trials, or births, in this case)
• $p = 0.5$ (the probability of each trial resulting in a girl)

Part (a): Mean

The mean of a binomial distribution is calculated as: $\mu = n \cdot p$

Plugging in the values: $\mu = 20 \cdot 0.5 = 10$

So, the mean number of girls is 10.

Part (b): Standard Deviation

The standard deviation for a binomial distribution is given by: $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$

Plugging in the values: $\sigma = \sqrt{20 \cdot 0.5 \cdot (1 - 0.5)} = \sqrt{20 \cdot 0.5 \cdot 0.5} = \sqrt{5} \approx 2.24$

So, the standard deviation of the number of girls is approximately 2.24.

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Let me know if you'd like more details or have any questions about these calculations.

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Here are five follow-up questions you might explore:

1. What would the mean and standard deviation be if the probability of having a girl was 0.6 instead?
2. How would these values change if there were 50 couples instead of 20?
3. What is the probability of having exactly 10 girls in a group of 20 births?
4. How does the standard deviation help in understanding the distribution of outcomes?
5. What is the probability of having more than 15 girls in a group of 20 births?

Tip: In a binomial distribution, if the probability remains the same, increasing the number of trials increases both the mean and the variability (standard deviation) of the outcome.