# Finding sample mean for normal distribution

#### Stardust71

##### New Member
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.

#### katxt

##### Well-Known Member
That's right. Then put in two scales - X is real life numbers, z is standard normal numbers.

#### katxt

##### Well-Known Member
Fine. It looks like you did it by trial and error. A more direct and accurate approach would be to find z that matches with 90% = 1.281552 (use the Excel formula =NORMSINV(0.9) ) and take that many SDs off your limit of 5. This is 2.437

#### obh

##### Well-Known Member
If you choose "p(X≤₁):" and 0.9
Mean:0, SD:2
http://www.statskingdom.com/normal.html

₁ = 2.563103, z1 = 1.281552.
P( X ≤ 2.563103 ) = 0.9.
P( X > 2.563103 ) = 0.1.
P(-2.563103 ≤ X ≤ 2.563103 ) = 0.8.
P(-3.289707 ≤ X ≤ 3.289707 ) = 0.9.
Probability density( 2.563103 ) = 0.0877492.

Then: 2.563103+Mean'=5 => Mean'=5-2.563103 = 2.4369

#### Stardust71

##### New Member
Thank you guys for taking time to explain. @katxt I used Norm inverse to arrive at 2.4 days.. and then used trial and error to figure out that the solution is less than or equal to 2.4 as it confused me. Thank you for helping out. @obh Thank you for the detailed explanation 