College STEM ISEN 3710 Engineering Statistics 1. A production manager has data o
ID: 2921881 • Letter: C
Question
College STEM ISEN 3710 Engineering Statistics 1. A production manager has data on three machines A, B, and C that indicates 9 out of 1000 parts from machine A, 16 out of 2000 parts from machine B, and 15 out of 1500 parts from machine C are defective. a.)what is the conditional probability of "a defect, qiven, it cane from machi b.) If an assembler that has received25, 35, and 40t of his - s Erom machines A, B, and C, respectively, what is the probability that, upon discovery of a defective part, it came from machine B, i.e., P(BID)2 that the average weight of parcels conforming to certairn outside dinensions is 90 lbs. probability of a conforming parcels being more than 100 1bs. with a variance of 40 lbs. What is the 3. A certain product is received in lots of 1000 units. The quality control dom from the lot and acceptance procedure is to choose 50 units at ran test them. If zero or one units in the sample are found defective the entire lot is rectified (bad accepted. What is the probability that a lot containing 5 defective parts will be accepted? Solve using the binomial distribution approximation to the hypergeometric distribution parts are replaced with known good parts) and 4. The number of calls per minute arriving at a telephone switchboard may be randomly distributed according to a poisson distribution. If an average of 500 calls per hour arrive at a switchboard, what is the probability that no more than 10 calls will arrive during any given minute?Explanation / Answer
Ans:
1)Given that
P(D/A)=9/1000=0.009
P(D/B)=16/2000=0.008
P(D/C)=15/1500=0.01
a)P(D/C)=15/1500=1/100=0.01
b)P(A)=0.25
P(B)=0.35
P(C)=0.4
So,
P(D)=0.009*0.25+0.008*0.35+0.01*0.4
P(D)=0.00905
Now,
P(B/D)=P(B and D)/P(D)=P(D/B)*P(B)/P(D)
=(0.008*0.35)/0.00905
= 0.0028/0.00905
=0.3094
2)mean=90
std deviation=sqrt(40)=6.32
z=(100-90)/6.32
z=10/6.32=1.58
P(z>1.58)=1-NORMSDIST(1.58)=0.0571
3)mean=500 calls per hour=500/60=8.33 calls per minute
P(k<=10)=POISSON(10,8.33,TRUE)=0.7818