Assume that the given procedure yields a binomial distribution with n trials and the probability of success for one trial is p. use the given values of n and p to find the mean mu and standard deviation sigma. also, use the range rule of thumb to find the minimum usual value mu minus 2 sigma and the maximum usual value mu plus 2 sigma. in an analysis of preliminary test results from a gender-selection method, 33 babies are born and it is assumed that 50% of babies are girls, so equals 33 and p equals 0.5.

Step-by-step explanation:

Hello!

The study variable is:

X: Number of female babies in a sample of 33 babies.

The variable has binomial distribution, symbolically:

X~Bi(n; p)

n= 33

p= 0.5

The mean of the Binomial variable is:

E(X)= n*p

E(X)= 33*0.5

E(X)= 16.5

The variance of the Binomial is:

V(X)= n*p*q

q= (1 - p) (If "p" is the probability of "success", "q" represents the probability of "failure")

V(X)= 33*0.5*0.5

V(X)= 8.25

Then the standard deviation is:

√V(X)= √8.25= 2.87

E(X) + 2*(√V(X))= 22.24

E(X) - 2*(√V(X))= 10.76

I hope it helps!

the 3rd option is right

step-by-step explanation: