Answer to Question #295549 in Statistics and Probability for Bhuwan

Question #295549

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.

1
Expert's answer
2022-02-10T04:18:49-0500

Solution;

On average,1 in 500 joints s defective;

Number of soldered joints,n=25

Probability that a joint is defective=1500\frac{1}{500}

Therefore;

Mean=no=25×1500=0.05Mean=no=25×\frac{1}{500}=0.05

Apply Poisson distribution;

λ=0.05\lambda=0.05

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

Hence;

P(r=0)=e0.05×0.0500!=e0.05P(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×e0.05=9512.29=10000×e^{-0.05}=9512.29


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