The problem in the image states:

"The random variable $x$ represents the number of girls in a family of three children. Assuming that boys and girls are equally likely, find the mean and standard deviation for the random variable $x$. Round to two decimal places."

Step 1: Define the problem.

• The number of trials (children) $n = 3$.
• The probability of having a girl $p = 0.5$ (since boys and girls are equally likely).
• The number of girls $x$ follows a binomial distribution: $x \sim Binomial(n = 3, p = 0.5)$.

Step 2: Mean of a binomial distribution.

The mean $\mu$ of a binomial distribution is given by:

$\mu = n \times p$

Substitute $n = 3$ and $p = 0.5$:

$\mu = 3 \times 0.5 = 1.5$

Step 3: Standard deviation of a binomial distribution.

The formula for the standard deviation $\sigma$ of a binomial distribution is:

$\sigma = \sqrt{n \times p \times (1 - p)}$

Substitute $n = 3$ and $p = 0.5$:

$\sigma = \sqrt{3 \times 0.5 \times (1 - 0.5)} = \sqrt{3 \times 0.5 \times 0.5} = \sqrt{0.75}$

$\sigma \approx 0.866$

Final Answer:

• Mean: $\mu = 1.50$
• Standard Deviation: $\sigma \approx 0.87$

Would you like further details or clarification on this solution?

Related Questions:

1. How would the mean and standard deviation change if the family had four children instead of three?
2. What is the probability of having exactly 2 girls in this scenario?
3. Can you calculate the variance for this binomial distribution?
4. How would the results change if the probability of having a girl was 0.6 instead of 0.5?
5. What is the probability of having no girls in a family of three children?

Tip:

In binomial distributions, the variance $\sigma^2$ is simply $n \times p \times (1 - p)$. This can help you quickly check or calculate deviations.