q = .35
thie looks like a binomial probability distribution problem.
if so, then, since there's a lot of calculations, i used excel to do the arithmetic.
what i get from excel is that the probabilkity of 120 or fewer of the individuals with the special characteristic is equal to 0.080469722.
this can also be solved using the normal approximation of the binomial formula.
if so, then:
np = .65 * 200 = 130 = the number of people out of 200 who have the characteristic.
s = sqrt(n * p * q) = sqrt(200 * .65 * .35) = 6.745368782 = the standard error of the distribution.
n is the number in the sample = 200
p is the proportion who have the characteristic = .65
q is the proportion who don't have the characteristic = 1-p = .35
z = (x-m)/s = (120.5-130)/s = -1.408373702
z is the z-score
m = the number of people from the sample who have the characteristic = n * p = 200 * .65 = 130
x = 120.5 = the number of people who you want to find the probbility being less than.
area to the left of that z-score of = -1.408373702 is equal to .0795102713
that's the probability of less thn or equal to 120 who have the characteristic.
the approximation is not too far off.
there's a calculqator online that does this for you.
that calculator can be found at https://homepage.divms.uiowa.edu/~mbognar/applets/binnormal.html
here's a reference on normal approximation of the binomial.
https://www.statisticshowto.com/probability-and-statistics/binomial-theorem/normal-approximation-to-the-binomial/
if you calculated the binomial distribution exactly, your solution is 0.080469722.
if you calculated the binomial distribution through the use of the normal approximation, your solution is .0795102713.
here are the results through the use of the online calculator.
