Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby. this method is designed to increase the likelihood that each baby will be a girl, but assume that the method has no effect, so the probability of a girl is 0.5. assume that the groups consist of 19 couples. complete parts (a) through (c) below. find the mean and the standard deviation for the numbers of girls in groups of 19 births.

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:

The standard deviation is:

In this problem:

The expected value is:

The standard deviation is:

The mean number of girls is 9.5, with a standard deviation of 2.18.

A similar problem is given at link

Step-by-step explanation:

For a binominal distribution you have the following formula for the mean and the standard distribution:

n: sample size

p: probability

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

i think it might be a but you might want a second opinion