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 43 couples. Complete parts​ (a) through​ (c) below.

a. Find the mean and the standard deviation for the numbers of girls in groups of 43 births.
The value of the mean is__ ​(Type an integer or a decimal. Do not​ round.)
The value of the standard deviation is __. ​(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 ___ girls or fewer are significantly low. ​(Round to one decimal place as​ needed.)
Values of ___ girls or greater are significantly high. (Round to one decimal place as​ needed.)
c. Is the result of 32 girls a result that is significantly​ high? What does it suggest about the effectiveness of the​ method?
The result is/is not significantly​ high, because 32 girls is greater than/equal to/less than ___ girls. A result of 32 girls would suggest that the method is not effective/is effective.​(Round to one decimal place as​ needed.)

Answer 1

A) Mean = 21.5

Standard deviation = 3.3

B) - Values of 14.9 girls or fewer are significantly low.

- Values of 28.1 girls or higher are significantly high

C) - The result is significantly​ high, because 32 girls is greater than 28.1 girls

-A result of 32 girls would suggest that the method is effective.

Explanation:

A) We are given;

Number of couples; n = 43

Probability it is a girl; p = 0.5

Formula for mean is;

Mean; x¯ = np

x¯ = 43 × 0.5

x¯ = 21.5

Formula for standard deviation is;

SD = √(np(1 - p))

SD = √(43 × 0.5 × (1 - 0.5))

SD = √(43 × 0.5²)

SD ≈ 3.3

B) range rule of thumb is that the values are usually within 2 standard deviations from the mean.

Thus, we have;

x¯ ± 2SD

This gives;

21.5 ± 2(3.3)

That is;

21.5 + 6.6 and 21.5 - 6.6

To get; 28.1 and 14.9

This means that;

- Values of 14.9 girls or fewer are significantly low.

- Values of 28.1 girls or higher are significantly high

C) If the result is 32,it doesn't fall in between 14.9 and 28.1. As a matter of fact it is higher than 28.1, therefore we can say it is significantly high

- A result of 32 girls would suggest that the method is effective.