James F. answered • 06/20/14
Tutor
5
(6)
Data Scientist and former Statistics Professor
The key to this problem is to recognize that density function as a Normal Probability Density function. Normal PDFs look like:
f(x) = 1/[√(2∏)σ]exp{-(1/(2σ^2)) (x-µ)^2}
From here, you can match the constants in your function to the parameters in the general form. Try to give it a shot. As a hint, take the (1/98) and pull out a (1/2). Compare what's leftover to the constant out front by the √(2∏). Does it match?
James F.
Sorry about that, I incorrectly typed the PDF above. The σ in the e^{} term should be σ^2.
So we want to take the 1/98 and pull at a 1/2, which leaves us with a 1/49. That 49 represents the σ^2. Now we can compare it to the term with the √(2∏), which is a 7.
7 = √49 ---> so σ = 7 and σ^2 = 49
Now for the µ term:
The general Normal PDF has (x-µ) in the e^{} term. Our specific one has (x+3.6). So our µ has to be -3.6.
J.T.
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06/21/14
Steve W.
06/21/14