Answer:
a.) 0.1028
b.) 0.6477
c.) 0.0388
d.) 3
e.) 2.55
Step-by-step explanation:
Forming a binomial Probability distribution
n = 20
Probability of success for Weld metal failure = 85%
Probability of success for base metal failure = 15%
We use the probabilit distribution formula of combination to solve the problem.
P(x=r) = nCr * p^r * q^n-r
a.) if exactly 5 are base metal failures, then p = 15 and our solution becomes:
P(x=5) = 20C5 * 0.15^5 * 0.85^15
P(x=5) = 0.1028
b.) probability that fewer than 4 are base metal failure= P(x=0) + P(x=1) + P(x=2) + P(x=3)
P(x=0) = 20C0 * 0.15^0 * 0.85^20 = 0.0388
P(x=1) = 20C1 * 0.15¹ * 0.85^19 = 0.1368
P(x=2) = 20C2 * 0.15² * 0.85^18 = 0.2293
P(x=3) = 20C3 * 0.15³ * 0.85^17 = 0.2428
Probability that fewer than 4 are base metal failures becomes: 0.038 + 0.1368 + 0.2293 + 0.2428 = 0.6477
c.) probability that none of them are results of base metal failure = P(x=0). As earlier calculated,
P(x=0) = 0.0388
d.) mean of base metal failures = np = 20*0.15 = 3
e.) standard deviation of base metal failures = √np(1-p)
=3 * (1 - 0.15) = 3 * 0.85
= 2.55
