Question 890018
Binomial distribution
 suppose that 50% of a population is left handed, find the probability that in a group of 50 individuals:
 a. at most 5 are left handed.
P(0<= x <=5) = binomcdf(50,0.5,5) = 0.000000002105

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 b. at least 5 are left handed.
P(6<= x <=50) = 1 - P(0<= x <=5) = 0.999999....
Cheers,
Stan H.