Questions: Refer to the accompanying table, which describes results from groups of 8 births from 6 different sets of parents. The random variable x represents the number of girls among 8 children. Find the mean and standard deviation for the number of girls in 8 births. The mean is μ = [ ] girl(s) (Round to one decimal place as needed.)

Solution Steps

The mean \( \mu \) of a binomial distribution can be calculated using the formula:

\[ \mu = n \cdot p \]

where:

• \( n = 8 \) (the number of trials),
• \( p \) is the probability of success (having a girl).

From the calculations, we find:

\[ \mu = 8 \cdot 0.5375 = 4.3 \]

The variance \( \sigma^2 \) of a binomial distribution is given by the formula:

\[ \sigma^2 = n \cdot p \cdot q \]

where \( q = 1 - p \). Thus, we have:

\[ \sigma^2 = 8 \cdot 0.5375 \cdot 0.4625 = 2.0 \]

The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{n \cdot p \cdot q} = \sqrt{2.0} \approx 1.4 \]

Final Answer