Let's begin by understanding the problem and breaking it down step by step.
We are given a scenario where a family has 5 children, and the probability of each child having freckles is 25%. We need to determine the mean ([tex]\(\mu\)[/tex]) and the standard deviation ([tex]\(\sigma\)[/tex]) of the number of children with freckles, denoted by [tex]\(X\)[/tex].
### 1. Identifying the Distribution:
Since each child either has or does not have freckles independently, and each has an equal probability, [tex]\(X\)[/tex] follows a Binomial Distribution. The parameters for a binomial distribution are:
- [tex]\(n\)[/tex] (number of trials) = 5 (number of children)
- [tex]\(p\)[/tex] (probability of success) = 0.25 (probability of having freckles)
### 2. Calculating the Mean ([tex]\(\mu\)[/tex]):
The mean of a Binomial Random Variable [tex]\(X\)[/tex] with parameters [tex]\(n\)[/tex] and [tex]\(p\)[/tex] is given by:
[tex]\[
\mu_x = n \cdot p
\][/tex]
Plug in the values:
[tex]\[
\mu_x = 5 \cdot 0.25 = 1.25
\][/tex]
### 3. Calculating the Standard Deviation ([tex]\(\sigma\)[/tex]):
The standard deviation of a Binomial Random Variable [tex]\(X\)[/tex] with parameters [tex]\(n\)[/tex] and [tex]\(p\)[/tex] is given by:
[tex]\[
\sigma_x = \sqrt{n \cdot p \cdot (1 - p)}
\][/tex]
Plug in the values:
[tex]\[
\sigma_x = \sqrt{5 \cdot 0.25 \cdot (1 - 0.25)}
\][/tex]
[tex]\[
\sigma_x = \sqrt{5 \cdot 0.25 \cdot 0.75}
\][/tex]
[tex]\[
\sigma_x = \sqrt{5 \cdot 0.1875}
\][/tex]
[tex]\[
\sigma_x = \sqrt{0.9375} \approx 0.968
\][/tex]
### 4. Comparing with Given Options:
By comparing our calculated values with the given answer options, we have:
- [tex]\(\mu_x = 1.25\)[/tex]
- [tex]\(\sigma_x \approx 0.97\)[/tex]
Hence, the correct answer is:
[tex]\[
\mu_x=1.25, \sigma_x=0.97
\][/tex]
So, the right option is:
[tex]\[
\boxed{\mu_x = 1.25, \sigma_x = 0.97}
\][/tex]
