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2 answers:
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Answer:
a) The expected number of calls in one hour is 30.
b) There is a 21.38% probability of three calls in five minutes.
c) There is an 8.2% probability of no calls in a five minute period.
Step-by-step explanation:
In problems that we only have the mean during a time period can be solved by the Poisson probability distribution.
Poisson probability distribution
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.
a. What is the expected number of calls in one hour?
Calls come in at the rate of one each two minutes. There are 60 minutes in one hour. This means that the expected number of calls in one hour is 30.
b. What is the probability of three calls in five minutes?
Calls come in at the rate of one each two minutes. So in five minutes, 2.5 calls are expected, which means that . We want to find P(X = 3).
There is a 21.38% probability of three calls in five minutes.
c. What is the probability of no calls in a five-minute period?
This is P(X = 0) with .
There is an 8.2% probability of no calls in a five minute period.
Answer:
Week Mass (gram)
(start) 80
1 88
2 96.8
3 106.48
4 117.128
Step-by-step explanation:
From the question we are told that
The mass of the kitten at the start week is m = 80 g
The rate at which the kitten grows is r = 0.10
Generally the mass of the kitten at the first week is mathematically represented as
=>
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Generally the mass of the kitten at the second week is mathematically represented as
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Generally the mass of the kitten at the third week is mathematically represented as
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Generally the mass of the kitten at the fourth week is mathematically represented as
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=>
Week Mass (gram)
(start) 80
1 88
2 96.8
3 106.48
4 117.128