I have this problem envolving normal distribution:
Consider 2 normal distributions that describe the daily waste of a family in city A and city B, respetively:
A~N(11.5;2.5^2)
B~N(9;3^2)
Mean and variance are in kilograms.
In 18.51% of the days, the families from city B produce waste that exceeds in over "x" kilograms the amount of waste produced by the families in city A.
Find x.
This is what i came up with:
P(B+x>A)=0.1851
P(B+x-A>0)=0.1851
with
B+x-A ~ N(-2.5+x;3.91^2)
using z-score:
1-P(Z<0)=0.1851
with
z=(0-(-2.5+x))/3.91
i got
P(Z<0)=0.8149
(2.5-x)/3.91=Φ^(-1)(0.8149)
which gave me
x=-1.019
with
Φ^(-1)(0.8149) =0.9 (aprox.)
I'm trying to solve this problem and this answer doesn't seem to be the correct one given the context of the problem. I don't know if it's correct to consider only 1 family from each city in my calculations given the problems asks for the total of families, but the problem doesn't give me the total amount of families in each city. I also assume working with one family versus working with all the families is just a matter of proportion that will give the same result. Correct me if my reasoning is wrong.
Where am I making the mistake(s) in this exercice?
I would have thought you should end up with $$x=(9-11.5) +\Phi^{-1}(1-0.1851) \sqrt{2.5^2 +3^2} \approx +1$$ and your initial error in the sign of $x$ is in trying to consider $\mathbb P(B+x>A)$ when the question is asking about $\mathbb P(B-A > x)=\mathbb P(B-x > A)$