SOLUTION: monthly sale of beer at a bar is believed to be approximately normally distributed with mean 2450 units and standard 400 units.to determine the level of orders and stock ,the manag

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Question 1197513: monthly sale of beer at a bar is believed to be approximately normally distributed with mean 2450 units and standard 400 units.to determine the level of orders and stock ,the management wants to find two value symmetrically on either side of mean,such that the probability that sales of beer during the month will be between the two values is 0.95 and 0.99 find the required value
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi  
 µ = 2450 and σ = 400   
the management wants to find two value symmetrically on either side of mean,
such that the probability that sales of beer during the month 
will be between the two values is 0.95 and 0.99 find the required value
If I am understanding Your Question as intended:
 
OR  
  
 95%  ⇒ z = 1.96
      x = 2450 ± 400*(1.96) 
      x = 2450 ± 784

99% ⇒ z = 2.576
      x = 2450 ± 400*(2.576) 
      x = 2450 ± 1030.4

 = CI	z = value
90%	z =1.645    
92%	z = 1.751
95%	z = 1.96   Invnorm(.975)  2-sided (.95 + .05/2 = .975 )
98%	z = 2.326  
99%	z = 2.576
Standard Normal Curve:

Wish You the Best in your Studies.

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