SOLUTION: Suppose that in a key punching of 80 column IBM cards, the arithmetic mean number of mistakes per card is 0.3. What percent of cards will have (i) no mistake, (ii) one mistake (iii
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-> SOLUTION: Suppose that in a key punching of 80 column IBM cards, the arithmetic mean number of mistakes per card is 0.3. What percent of cards will have (i) no mistake, (ii) one mistake (iii
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Question 1058198: Suppose that in a key punching of 80 column IBM cards, the arithmetic mean number of mistakes per card is 0.3. What percent of cards will have (i) no mistake, (ii) one mistake (iii) two mistakes? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! random, proportional to number, could be infinite, discrete
Poisson distribution with parameter 0.3
e^(-lambda)*lambda^x/x!
For 0, it is e^(-0.3)=0.7408
for 1 it is 0.7408(0.3)=0.2222
for 2 it is 0.7408*(0.3^2)/2=0.0334
