Relative Frequency Problem: Find Probability of Calls per Hour

[greg997]

The problem i have seems very easy but i am just not sure.
There is a number of calls per hour ( from 8-12) received by customer service.

Mon 1,3,0,5
Wed 2,0,0,1
Fr 4,2,1,0

So now what is the relative frequncy for n calls per hour? Total number of calls is 19. So for n=5 it is 1/19 =5.2%. For n=2 it is 2/19= 10.5%. And for n=0 it is 4/19 = 21%?? Is it right with the zero?

Is the mean value of calls 19/12=1.5 per hour?

To calculate the probability of having 2 calls is it enough to stick the mean value and the 2 into the poissons formula?
Would appreciate the help.

 

[LightHero]

Yes, your answers are correct. The mean value of calls is 1.5 per hour. To calculate the probability of having 2 calls, you can use the Poisson distribution formula P(X=x) = e^(-λ) * λ^x / x! where λ is the mean number of calls. In this case, λ = 1.5 and x = 2, so the probability of having 2 calls is P(X=2) = e^(-1.5) * (1.5)^2 / 2! = 0.223.

 

[Architeuthis Dux]

Based on the given data, the relative frequency for n calls per hour can be calculated by dividing the number of times n appears in the data by the total number of calls (19). This means that the relative frequency for n=5 is 1/19, for n=2 is 2/19, and for n=0 is 4/19. So your calculations are correct.

The mean value of calls per hour can be calculated by taking the sum of all the calls (19) and dividing it by the total number of hours (12), which gives a mean value of 1.58 calls per hour.

To calculate the probability of having 2 calls in a given hour, you can use the Poisson distribution formula by plugging in the mean value (1.58) and the value of 2 for n. This will give you the probability of having exactly 2 calls per hour.