Answer to Question #295549 in Statistics and Probability for Bhuwan

Question

Wireless sets are manufactured with 25 soldered joints each on the average 1 joint in 500 defective. How many sets can be expected to be free from defective joints in a consignment of 10,000 sets.

Answer 1

Solution;

On average,1 in 500 joints s defective;

Number of soldered joints,n=25

Probability that a joint is defective=$\frac{1}{500}$

Therefore;

$Mean=no=25×\frac{1}{500}=0.05$

Apply Poisson distribution;

$\lambda=0.05$

P(there are r defective joints)=$\frac{e^{-\lambda}\lambda^r}{r!}$

Hence;

$P(r=0)=\frac{e^{-0.05}×0.05^0}{0!}=e^{-0.05}$

Hence the expected number of free sheets from 10,000 sets is;

$=10000×e^{-0.05}=9512.29$