Question

in a family of five children, there is a 25% chance that each individual child has freckles. children inherit freckles independently of one another. let ( x ) represent the number of children in this family with freckles. what are the mean and standard deviation of ( x )?
( mu_{x}=1.25, sigma_{x}=0.97 )
( mu_{x}=1.25, sigma_{x}=0.94 )
( mu_{x}=1.5, sigma_{x}=0.19 )
( mu_{x}=1.5, sigma_{x}=0.57 )

Explanation:

Step1: Identify the distribution

This is a binomial distribution problem. The binomial distribution formula for the mean \(\mu = np\) and the standard deviation \(\sigma=\sqrt{np(1 - p)}\), where \(n\) is the number of trials and \(p\) is the probability of success in a single trial. Here, \(n = 5\) (number of children) and \(p=0.25\) (probability of a child having freckles).

Step2: Calculate the mean

Using the formula \(\mu=np\), substitute \(n = 5\) and \(p = 0.25\).

$$ \mu=5\times0.25=1.25 $$

Step3: Calculate the standard deviation

First, find \(1 - p=1 - 0.25 = 0.75\). Then use the formula \(\sigma=\sqrt{np(1 - p)}\). Substitute \(n = 5\), \(p = 0.25\), and \(1 - p=0.75\)

$$ \sigma=\sqrt{5\times0.25\times0.75}=\sqrt{0.9375}\approx0.97 $$

Answer:

\(\mu_{x}=1.25,\sigma_{x}\approx0.97\) (corresponding to the first option \(\mu_{x}=1.25,\sigma_{x}=0.97\))