In a Poisson probability problem the rate of errors is one e
In a Poisson probability problem, the rate of errors is one every two hours. To find the probability of three defects in four hours, which lambda (mean) and x variable values should be used?
? = 2, x = 3
?= 3, x = 2
?= 6, x =2
? = 1, x = 2
?= 3, x =6
| ? = 2, x = 3 | ||
| ?= 3, x = 2 | ||
| ?= 6, x =2 | ||
| ? = 1, x = 2 | ||
| ?= 3, x =6 |
Solution
Answer: ? = 2, x = 3
