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

3 0

3 years ago

Please help it’s pretty easy and I’ll reward brainiest

Julli [10]

A, hope this helps!!

8 0

3 years ago