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 42 couples. Complete parts​ (a) through​ (c) below.
a. Find the mean and the standard deviation for the numbers of girls in groups of 48 births.
b. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.
c. Is the result of 42 girls a result that is significantly​ high? What does it suggest about the effectiveness of the​ method?

Answer 1

Answer:

a) $\mu = 24,\sigma = 3.46$

b) Significantly low:

$x < 17.08$

Significantly high:

$x > 30.92$

c) 42 girls are significantly high.                                      

Step-by-step explanation:

We are given the following in the question:

$p = 0.5$

a) mean and the standard deviation for the numbers of girls in groups of 48 births

$\mu = np = 48(0.5) = 24\\\sigma = \sqrt{np(1-p)} = \sqrt{48(0.5)(1-0.5)} = 3.46$

b) Range rule of thumb

Significantly low: According to this rule the observations lying below two standard deviation of mean is considered significantly low.

$x = \mu - 2\sigma\\x = 24 - 2(3.46) = 17.08$

Significantly high: According to this rule the observations lying above two standard deviation of mean is considered significantly high.

$x = \mu + 2\sigma\\x = 24 + 2(3.46) = 30.92$

c) Significance of 42 girls

$42 > \mu + 2\sigma$

Since 42 is greater than 30.92, 42 girls are significantly high. Thus, the method is not significant.