Answer:
a) 0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.
b) 0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.
c) 0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.
d) 0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.
e) 0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
Poisson distributed, with a mean of 0.6 trips per year
This means that , in which n is the number of years.
a.The family did not make a trip to an amusement park last year.
This is P(X = 0) when n = 1, so .
0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.
b.The family took exactly one trip to an amusement park last year.
This is P(X = 1) when n = 1, so .
0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.
c.The family took two or more trips to amusement parks last year.
Either the family took less than two trips, or it took two or more trips. So
We want
In which
0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.
d.The family took three or fewer trips to amusement parks over a three-year period.
Three years, so .
This is
0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.
e.The family took exactly four trips to amusement parks during a six-year period.
Six years, so .
This is P(X = 4). So
0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.
