Question

Assume that different groups of couples use the XSORT method of gender selection and each couple gives birth to one baby. The XSORT method is designed to increase the likelihood that a baby will be a girl, but assume that the method has no effect, so the probability of a girl is 0.5. Now consider a group consisting of 36 couples. a.) Find the mean and standard deviation for the numbers of girls in groups of 36 births

Answer 1

Answer:

Mean and Standard deviation for the numbers of girls in groups of 36 births are 18 and 3 respectively.

Step-by-step explanation:

We are given that he X SORT method is designed to increase the likelihood that a baby will be a girl, but assume that the method has no effect, so the probability of a girl is 0.5.

Now consider a group consisting of 36 couples.

The above situation can be represented through binomial distribution;

$P(X=r)=\binom{n}{r} \times p^{r} \times (1-p)^{n-r} ;x=0,1,2,3,.....$

where, n = number of trials (samples) taken = 36 couples

            r = number of success

            p = probability of success which in our question is probability

                   of a girl, i.e.; p = 0.5

Let X = Numbers of girls in groups of 36 births

So, X ~ Binom(n = 36, p = 0.5)

Now, mean for the numbers of girls in groups of 36 births is given by;

         Mean, E(X) = $n \times p$ = $36 \times 0.5$ = 18

Also, standard deviation for the numbers of girls in groups of 36 births is given by;

        Standard deviation, S.D.(X) =  $\sqrt{n\times p\times (1-p)}$

                                                      =  $\sqrt{36\times 0.5\times (1-0.5)}$

                                                      =  3