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,.....](https://tex.z-dn.net/?f=P%28X%3Dr%29%3D%5Cbinom%7Bn%7D%7Br%7D%20%5Ctimes%20p%5E%7Br%7D%20%5Ctimes%20%281-p%29%5E%7Bn-r%7D%20%3Bx%3D0%2C1%2C2%2C3%2C.....)
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
<em><u>Let X = Numbers of girls in groups of 36 births </u></em>
So, X ~ Binom(n = 36, p = 0.5)
Now, mean for the numbers of girls in groups of 36 births is given by;
<u>Mean</u>, E(X) =
=
= 18
Also, standard deviation for the numbers of girls in groups of 36 births is given by;
<u>Standard deviation</u>, S.D.(X) =
=
= 3