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

4/6 = 2/3

so it should take 3/3 hours per man. so 9 hours is the answer 3*3 =9

step-by-step explanation:

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %