Given a Poisson random variable x, where the average number of times an event occurs in a certain period of time or space is 1.5, then P(x = 2) is:
a. |
0.5 |
|
b. |
0.5020 |
|
c. |
0.2510 |
|
d. |
0.2231 |
|
e. |
0.1116 |
A | B | C | D | E | F | G | H | I | J | K |
2 | ||||||||||
3 | As per Poisson distribution, the probability of observing k events over time is | |||||||||
4 | P(X=k) = (λke-λ)/k! | |||||||||
5 | ||||||||||
6 | Where λ is the average number of times the event occurs over the time period. | |||||||||
7 | ||||||||||
8 | Since the average number of times an event occurs in a certain period of time is 1.5 | |||||||||
9 | therefore | |||||||||
10 | λ | 1.5 | ||||||||
11 | Thus the probability that the even will occur twice in the given period of time can be calculated as follows: | |||||||||
12 | P(X=2) | = (λ2e-λ)/2! | ||||||||
13 | 0.2510 | =POISSON.DIST(2,D10,FALSE) | ||||||||
14 | ||||||||||
15 | Hence the probability is | 0.2510 | ||||||||
16 | Thus the option (c) is correct. | |||||||||
17 |
Given a Poisson random variable x, where the average number of times an event occurs in...
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