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.

Answer :

Step 1 of 6:

Given, in a sample of 20 weld failures, 85% of them occur in the weld metal itself, and the remaining 15% occur in the base metal.

Let X follows binomial distribution with probability mass function is

P(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.

Where,

n = sample size

= 20

x = random variable

p = probability of success

= 15%

= 0.15

q = 1 - p (probability of failure)

Mean of the binomial distribution is

= np

Variance of the binomial distribution is

= npq

Step 2 of 6:

The claim is to find the probability that exactly five of them are base metal failuresWe have to find P(X=5)

P(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.

Where, n = 20, p = 0.15 and x = 5

P(x = 5) = (1 - 0.15

= (15504) (0.0000759) (0.087354)

= 0.1028

Hence, the probability that exactly five of them are base metal failures is 0.1028.

Step 3 of 6:

The claim is to find the probability that fewer than four of them are base metal failuresWe have to find P(X < 4) = P(x = 0) + P( x = 1) + P(x = 2) + P(x = 3)

P(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.

Where, n = 20, p = 0.15 and x = 0, 1, 2 and 3

P(x < 4) = (1 - 0.15+ (1 - 0.15+

(1 - 0.15+ (1 - 0.15

= (1) (1) (0.03876) + (20) (0.15) (0.0456) +

(190) (0.0225) (0.053646) + (1140) ( 0.003375) (0.0631134)

= 0.03876 + 0.1368 + 0.2293 + 0.2428

= 0.6477

Hence, the probability that fewer than four of them are base metal failures 0.6477