Answer:
[tex]\mu = 9.5 \\\sigma= 2.178[/tex]
Step-by-step explanation:
For a binominal distribution you have the following formula for the mean and the standard distribution:
[tex]\mu = n*p \\\sigma = \sqrt{n*p*(1-p)}[/tex]
n: sample size
p: probability
[tex]\mu = 19 * 0.5 = 9.5\\\sigma = \sqrt{19*0.5*(1-0.5)} =2.179[/tex]
Using the binomial distribution, we have that:
The mean number of girls is 9.5, with a standard deviation of 2.18.
For each baby, there are only two possible outcomes. Either it is a girl, or it is not. The probability of a baby being a girl is independent of any other baby, which means that the binomial distribution is used to solve this question.
The binomial distribution is the probability of x successes on n trials, with p probability.
The expected value is:
[tex]E(X) = np[/tex]
The standard deviation is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In this problem:
The expected value is:
[tex]E(X) = np = 19(0.5) = 9.5[/tex]
The standard deviation is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{19(0.5)(0.5)} = 2.18[/tex]
The mean number of girls is 9.5, with a standard deviation of 2.18.
A similar problem is given at https://brainly.com/question/24261244
