Question

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 30 couples. complete parts (a) through (c) below.
a. find the mean and the standard deviation for the numbers of girls in groups of 30 births.
the value of the mean is μ = 15.
(type an integer or a decimal. do not round.)
the value of the standard deviation is σ = 2.7.
(round to one decimal place as needed.)
b. use the range rule of thumb to find the values separating results that are significantly low or significantly high.
values of 9.6 girls or fewer are significantly low.
(round to one decimal place as needed.)
values of 20.4 girls or greater are significantly high.
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean and standard deviation

For a binomial distribution \(n = 30\) (number of trials, here number of births in a group), \(p=0.5\) (probability of having a girl).
The mean formula for a binomial distribution is \(\mu=np\). Substituting \(n = 30\) and \(p = 0.5\), we get \(\mu=30\times0.5=15\).
The standard deviation formula for a binomial distribution is \(\sigma=\sqrt{np(1 - p)}\). Substituting \(n = 30\) and \(p = 0.5\), we have \(\sigma=\sqrt{30\times0.5\times(1 - 0.5)}=\sqrt{30\times0.5\times0.5}=\sqrt{7.5}\approx2.7\).

Step2: Use the range rule of thumb

The range rule of thumb states that values are significantly low if \(x\leq\mu - 2\sigma\), and significantly high if \(x\geq\mu+2\sigma\).
First, calculate \(\mu - 2\sigma\): \(15-2\times2.7=15 - 5.4 = 9.6\).
Then, calculate \(\mu+2\sigma\): \(15 + 2\times2.7=15+5.4=20.4\).

Answer:

a. Mean \(\mu = 15\), standard deviation \(\sigma\approx2.7\)
b. Significantly low: \(9.6\) girls or fewer; Significantly high: \(20.4\) girls or greater.