Answer:
The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.
Explanation:
With the weekly average we can estimate the daily average for customers, assuming 7 days a week:
![M=756/7=108](https://img.qammunity.org/2021/formulas/mathematics/college/guct3gh3v4poxvyx6108lleeybefww2k5p.png)
We can model this situation with a Poisson distribution, with parameter λ=108. But because the number of events is large, we use the normal aproximation:
![P(\lambda)\approx N(\lambda,\lambda)](https://img.qammunity.org/2021/formulas/mathematics/college/vdevhqd3oyuhp1vvuljrvg2vcu2mejd9cm.png)
Then we can calculate the z value for x=100:
![z=(x-\mu)/(\sigma)=(100-108)/(√(108))=(-8)/(10.4) =-0.77](https://img.qammunity.org/2021/formulas/mathematics/college/jqwuauzbqqgz4ytu0zj8yboau0jrybait5.png)
Now we calculate the probability of x>100 as:
![P(x>100)=P(z>-0.77)=0.78](https://img.qammunity.org/2021/formulas/mathematics/college/r79n8m7e408coziw4wqtmd5ssvglhrwsx5.png)
The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.