## Of all the weld failures in a certain assembly, 85% of them occur in the weld metal itself, and the remaining 15% occur in the base metal. A

Question

Of all the weld failures in a certain assembly, 85% of them occur in the weld metal itself, and the remaining 15% occur in the base metal. A sample of 20 weld failures is examined.a.  What is the probability that exactly five of them are base metal failures?b.  What is the probability that fewer than four of them are base metal failures?c.  What is the probability that none of them are base metal failures?d.  Find the mean number of base metal failures,e.  Find the standard deviation of the number of base metal failures.

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2 months 2021-10-15T01:50:13+00:00 1 Answer 0 views 0

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