The probability that an individual is left-handed is 0.12. in a class of 39 students, what is the probability of finding five left-handers

We are given:

p = probability = 0.12

n = total students = 39

x = left handers = 5

u = mean = p * n = 4.68

σ = standard dev = √ (n*p * (1-p)) = √ (39 * 0.12 * 0.88) = 2.03

Calculating for the z score:

z = (x - u) / σ

z = (5 - 4.68) / 2.03

z = 0.1576 = 0.16

Using the standard tables for z, the p value is:

p value = 0.5636 = 56.36%

Hence there is a 56.36% chance.