of all the weld failures in a certain assembly, 85% of | studysoup
?problem 7e


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?
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b. what is the probability that fewer than four of them are base metal failures?
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c. what is the probability that none of them are base metal failures?
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d. find the mean number of base metal failures.
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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


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step 2 of 6:




1. the claim is to find the probability that exactly five of them are base metal failures


we 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.


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step 3 of 6:




1. the claim is to find the probability that fewer than four of them are base metal failures


we 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


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step 4 of 6:




c) the claim is to find the probability that none of them are base metal failures.


we have to find p(x=0) 


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


 where, n = 20, p = 0.15 and x = 0


p(x = 0) = (1 - 0.15


 = (1) (1) (0.03876) 


 = 0.03876


hence, the probability that none of them are base metal failures is 0.03876


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step 5 of 6:




d) the claim is to find the mean number of base metal failures..


 mean of the binomial distribution is = np


where, n = 20, and p = 0.15


substitute these values in 


then, = (20) (0.15)


 = 3


hence, the mean number of of base metal failures is 3.


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step 6 of 6:




e) the claim is to find the standard deviation of the number of base metal failures..


standard deviation of the binomial distribution is 


where, n = 20, p = 0.15 and q = 1 - p


 = 1 - 0.15


 = 0.85


substitute these values in 


then, 


 


 = 1.596


hence, the standard deviation of the number of base metal failures is 1.596.


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