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 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 47 couples. complete parts (a) through (c) below.
a. find the mean and the standard deviation for the numbers of girls in groups of 47 births.
the value of the mean is $mu=square$
(type an integer or a decimal. do not round.)
the value of the standard deviation is $sigma=square$
(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 $square$ girls or lower are significantly low
(round to one decimal place as needed.)
values of $square$ girls or greater are significantly high
(round to one decimal place as needed.)
c. is the result of 40 girls a result that is significantly high? what does it suggest about the effectiveness of the method?
the result $square$ significantly high, because 40 girls is $square$ girls. a result of 40 girls would suggest that the method $square$
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean

The mean $\mu$ of a binomial distribution is given by $\mu = np$, where $n$ is the number of trials and $p$ is the probability of success on each trial. Here, $n = 47$ (number of couples) and $p=0.5$ (probability of having a girl). So, $\mu = 47\times0.5=23.5$.

Step2: Calculate the standard - deviation

The standard deviation $\sigma$ of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$. Substitute $n = 47$ and $p = 0.5$ into the formula: $\sigma=\sqrt{47\times0.5\times(1 - 0.5)}=\sqrt{47\times0.5\times0.5}=\sqrt{11.75}\approx3.4$.

Step3: Use the range - rule of thumb for significant values

The range - rule of thumb states that significant low values are $\mu-2\sigma$ and significant high values are $\mu + 2\sigma$.
Significant low values: $\mu-2\sigma=23.5-2\times3.4=23.5 - 6.8 = 16.7$.
Significant high values: $\mu+2\sigma=23.5 + 2\times3.4=23.5+6.8 = 30.3$.

Step4: Determine if 40 girls is a significant result

Since $40>30.3$, the result of 40 girls is significantly high. This suggests that the method may be effective in increasing the likelihood of having a girl.

Answer:

a. The value of the mean is $\mu = 23.5$
The value of the standard deviation is $\sigma\approx3.4$
b. Values of $16.7$ girls or lower are significantly low
Values of $30.3$ girls or greater are significantly high
c. The result is significantly high, because 40 girls is $>30.3$ girls. A result of 40 girls would suggest that the method may be effective.