Step-by-step explanation:
a. The random variable is a binomial distribution.
b. the sample space, X = {0, 1, 2, 3, 4, 5}
the pmf
we solve for this using
nCx * P^x * (1-p)^n-x
n = 5
p = 0.3
for x = 0
5C0 * 0.3⁰(1-0.3)^5-0
= 1 * 1* 0.7⁵
= 0.16807
for x = 1
5C1*0.3¹(1-0.3)^5-1
= 5*0.3(0.7)⁴
= 5x0.3x0.2401
= 0.36015
for x = 2
5C2 * 0.3² * (1-0.3) ^5-2
= 0.30870
for x = 3
5C3 * 0.3³ * (1-0.3) ^ 5-3
= 10 * 0.027 * 0.7²
= 0.1323
for x = 4
5C4 * 0.3⁴ (1-0.3) ^ 5-4
= 5 * 0.0081 * 0.7
= 0.02835
for x = 5
5C5 *0.3⁵ (1- 0.3) ^5-5
= 1*0.00243*0,7⁰
= 0.00243
c. E[X] = N*p = 5*0.3 = 1.5
var[X] = np(1-p) = 5*0.3*0.7 = 1.05
d. 20/9 = 2.222
so we have that if x is greater than or equal to 3, cost will exceed 20
p(x=3) + p(x=4) + p(x=5)
= 0.1323 + 0.02835 + 0.00243
probability = 0.16308
E[C] = 1.5 * 9 = 13.5
VAR[C] = 1.05 * 9 = 9.45