Finding sample mean for normal distribution

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Stardust71

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Hello friends,

I require help on the below question on normal distribution.

"An e-commerce company tries to settle the customer complaints on refund within 5 days. If a refund is not processed within 5 days, then the company has to provide coupons for Rs.500 to the customer. If the company targets that at least 90% of the refund related complaints should be processed within 5 days, then what should be mean processing time. The standard deviation is 2 days."

I'm confused whether the given 90% is P value as 0.9 or to be taken as alpha 0.1. Please help with the solution.
Thanks in advance.
 
Stardust71 said:
"An e-commerce company tries to settle the customer complaints on refund within 5 days. If a refund is not processed within 5 days, then the company has to provide coupons for Rs.500 to the customer. If the company targets that at least 90% of the refund related complaints should be processed within 5 days, then what should be mean processing time. The standard deviation is 2 days."

I'm confused whether the given 90% is P value as 0.9 or to be taken as alpha 0.1. Please help with the solution.
Thanks in advance.
Click to expand...
I wouldn't really call it either, as this is not a hypothesis test. Maybe you're taught to use those terms even here; but as I see it, you need to think about the meaning of what you are doing, not just follow some routine based on the name given to a quantity.

I would just draw a normal curve and mark what area has to be 90%. Then find z for that probability.

Which side of the line did you shade in?
 
Thank you. Plotting it really helped clear my mind.

For exactly 90% of the coverage, my mean comes around 2.4 days.
Since the question is about atleast 90% coverage, I believe my population mean should be less than or equal to 2.4 days.

Is my understanding correct?

1594956858284.png
 
That looks about right.

Did you get 2.4 by trial and error, or did you have a calculation that led to the answer? (I have an answer that's a little more precise.)
 
I used Norm inverse to arrive at 2.4 days.. But used trial and error to figure out that the solution is less than or equal to 2.4..
 
Stardust71 said:
I used Norm inverse to arrive at 2.4 days.. But used trial and error to figure out that the solution is less than or equal to 2.4..
Click to expand...

I'm not quite sure of the difference.

I found z from the 90% (I'd call that step norm inverse), then solved algebraically to find mu, which came to 2.44. I'm hoping you just rounded more than I did.
 
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