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 19 couples.
a. Find the mean and the standard deviation for the numbers of girls in groups of 19 births.
b. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.

Answer 1

Answer:

a. Mean: 9.5.

Standard deviation: 2.18.

b. The critical value for signficantly low girl births is 5.14.

The critical value for signficantly high girl births is 13.86.

Step-by-step explanation:

a) If we take a sample of 19 births, all with individual probability of 0.5, we have a random binomial variable with n=19 and p=0.5.

This random variable will have mean and standard deviation calculated as:

$\mu=n\cdot p=19\cdot0.5=9.5\\\\\sigma=\sqrt{np(1-p)}=\sqrt{19\cdot0.5\cdpot0.5}=\sqrt{4.75}=2.18$

b) Range rule of thumb: The standard deviation is approximately a quarter of the range, difference between the maximum and minimum value.

The range rule of thumb tells us that the expected minimum and maximum value are 2 standards deviation below or above, respectively, from the mean.

Then, the extreme expected values are:

$Min=\mu-2\sigma=9.5-2\cdot2.18=9.5-4.36=5.14\\\\Max=\mu+2\sigma=9.5+2\cdot2.18=9.5+4.36=13.86$

The critical value for signficantly low girl births is 5.14.

The critical value for signficantly high girl births is 13.86.