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Sagot :
Answer:
0.18203 = 18.203% probability that exactly four complaints will be received during the next eight hours.
Step-by-step explanation:
We have the mean during a time-period, which means that the Poisson distribution is used to solve this question.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
A service center receives an average of 0.6 customer complaints per hour.
This means that [tex]\mu = 0.6h[/tex], in which h is the number of hours.
Determine the probability that exactly four complaints will be received during the next eight hours.
8 hours means that [tex]h = 8, \mu = 0.6(8) = 4.8[/tex].
The probability is P(X = 4).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 4) = \frac{e^{-4.8}*4.8^{4}}{(4)!} = 0.18203[/tex]
0.18203 = 18.203% probability that exactly four complaints will be received during the next eight hours.
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