Answer:
The probability that more than 4 calls will be received during the noon hour, on some particular Thursday is 0.9113.
Step-by-step explanation:
The random variable X can be defined as the number of incoming phone calls during the noon hour on Thursdays.
The random variable X describes a finite number of occurrences of an event in a fixed time interval.
The random variable X follows a Poisson distribution with parameter λ = 8.2.
The probability mass function of X is:
[tex]P(X=x)=\frac{e^{-8.2}(8.2)^{x}}{x!};x=0,1,2,3...[/tex]
Compute the probability of more than 4 calls as follows:
[tex]P(X>4)=1-P(X\leq 4)[/tex]
[tex]=1-\sum\limits^{4}_{x=0}{\frac{e^{-8.2}(8.2)^{x}}{x!}}\\\\=1-[0.00027+0.00225+0.00923+0.02524+0.05174]\\\\=1-0.08873\\\\=0.91127\\\\\approx 0.9113[/tex]
Thus, the probability that more than 4 calls will be received during the noon hour, on some particular Thursday is 0.9113.
