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Expected Value and Variance?
A certain typing agency employs two typists. The average number of errors per article is 3.8 when typed by the first typist and 2.3 when typed by the second. If your article is equally likely to be typed by either typist, find the probability that it will have no errors.
So I get how each is .5 probability of getting either typist.
So that's E(x) of 3.8 (.5) + 2.3(.5).
But how do I find the probability of 0 errors?
2 Answers
- BillyLv 57 years ago
Hi there :)
To be blunt, PrivateBanker's solution is close but wrong; 3.8 and 2.3 are expected values; there is no reason to consider ".038" and ".023" respectively.
Personally, I think there isn't enough information. But if I had to make an informed guess, I would say that 3.8 (and 2.3) is the mean of a binomial random variable with n = 1 (article). Since the binomial mean = n*p, we have that the probabilities are about 0.2632 and 0.4348.
Hence, their complements are 0.7368 abd 0.5652.
Then, under law of total probability,
P(no errors) =
P(no errors and typist 1) + P(no errors and typist 2)
P(no errors|typist1)*P(typist 1) + P(no errors|typist2)*P(typist 2) =
0.7368*.5 + 0.5652*.5 = 0.651
- PrivateBankerLv 77 years ago
The probability of no errors is (1 - probability of errors) for each typist.
0.50(1 - 0.038) + 0.50(1 - 0.023)
= 0.481 + 0.4885
= 0.9695, or 96.95%
