Question

Suppose that you are in a class of 32 students and it is assumed that approximately 18% of the population is left-handed. (Give your answers correct to three decimal places.)

(a) Compute the probability that exactly five students are left-handed.

(b) Compute the probability that at most four students are left-handed.

(c) Compute the probability that at least six students are left-handed.

Answer 1

solution:-
given that p = 0.18 , n = 32
formula
p(X = x) = ncr * p^r * q^(n-r)
(a) p(x = 5) = 32c5 * (0.18)^5 * (0.82)^27
= 0.179

(b)p(x ≤ 4) = p(x = 0) + p(x = 1) +p(x = 2) +p(x = 3) +p(x = 4)
= 32c0 * (0.18)^0 * (0.82)^32 + 32c1 * (0.18)^1 * (0.82)^31 +32c2 * (0.18)^2 * (0.82)^30 + 32c3 * (0.18)^3 * (0.82)^29 + 32c4 * (0.18)^4 * (0.82)^28
= 0.0017 + 0.0123 + 0.0417 + 0.0916+0.1458
= 0.293
(c) p(x ≥ 6) = 1 - ( p(x = 0) + p(x = 1) +p(x = 2) +p(x = 3) +p(x = 4) + p(x= 5))
= 1 - (32c0 * (0.18)^0 * (0.82)^32 + 32c1 * (0.18)^1 * (0.82)^31 +32c2 * (0.18)^2 * (0.82)^30 + 32c3 * (0.18)^3 * (0.82)^29 + 32c4 * (0.18)^4 * (0.82)^28 + 32c5 * (0.18)^5 * (0.82)^27)
= 1 - (0.0017 + 0.0123 + 0.0417 + 0.0916+0.1458 + 0.1792)
   = 1 - (0.472)
= 0.528