Lake Aberdeen has a mean phosphorus conc. of 14.6 mg/L and standard deviation of 5.8 mg/L. Random day selected. Q: What is the probability that the total phosphorus differs from the mean by more than 5mg/L? Can you explain how to do this problem? Thank-you Solution Normal Distribution Mean ( u ) =14.6 Standard Devaition ( sd )=5.8 Normal Distribution = Z= X- u / sd ~ N(0,1) P(X > 19.6) = x-U/5.8 = 5/5.8 = 0.8621 = P ( Z >0.8621) From Standard Normal Table = 0.19432.

1. Lake Aberdeen has a mean phosphorus conc. of 14.6 mg/L and standard deviation of 5.8 mg/L.
Random day selected.
Q: What is the probability that the total phosphorus differs from the mean by more than 5mg/L?
Can you explain how to do this problem? Thank-you
Solution
Normal Distribution
Mean ( u ) =14.6
Standard Devaition ( sd )=5.8
Normal Distribution = Z= X- u / sd ~ N(0,1)
P(X > 19.6) = x-U/5.8
= 5/5.8
= 0.8621
= P ( Z >0.8621) From Standard Normal Table
= 0.19432