Answer:
2.35 calls
Step-by-step explanation:
The presented scenario can be modeled by a Poisson distribution with an average number of calls (μ) of 5.5 during the noon hour on Mondays.
Therefore, the standard deviation for the number of calls received, X, is given by:
[tex]\sigma =\sqrt \mu\\\sigma =\sqrt5.5\\\sigma =2.35\ calls[/tex]
The standard deviation of X is 2.35 calls.
Answer:
Standard deviation of X = 2.3452 calls
Step-by-step explanation:
This is a Poisson distribution problem with an average (μ) of 5.5 number of incoming calls during the noon hour on Mondays.
Therefore, the standard deviation for the number of calls received, X, is represented by:
a. σ = √µ
Thus;
X ~ P(5.5); µ = 5.5; σ = √5.5 ≈ 2.3452 calls
