Question
Smith is a weld inspector at a shipyard. He knows from keeping track of good and substandard welds that for the afternoon shift 10% of all welds done will be substandard. More than 100,000 welds were completed during one afternoon shift. If Smith randomly selects a sample of 144 welds then what is the probability that he will find 18 or more substandard welds? Round to 4 decimal places
Answer
Smith us a weld inspector at a shipyard. He knows from keeping track of good and substandard welds that for the afternoon shift 10℅ of all welds done will be substandard.
=> Given data :-
- Probability of finding a substandard weld (P) = 10℅ => P = 0.1
- Sample size (n) = 144
=> According to the Binomial Distribution:
* The average number of welds (m)
= m = n × p
= 144 × 0.1
=> m = 14.4
* The standard Deviation of weld = √m
= √14.4
=> sd = 3.7947
=> By using Normal Distribution, Z value corresponding to 18 is to be calculated..
=> Z = (X – mean)/standard deviation
= (18 – 14.4)/3.7947
= 3.6/3.7947
=> Z = 0.9486
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