Conditional normal probability, repost

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Homework Help Overview

The problem involves calculating the probability that one family will receive at least twice as much sunshine as another family, given that the number of hours of sunshine follows a normal distribution. The context includes two families spending different durations at a resort, with specific expectations and standard deviations for sunshine hours.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the distribution of sunshine hours for each family and consider the implications of comparing these distributions. There is an exploration of the relationship between the random variables A and B, particularly through the ratio Z = A/B and the suggestion to examine A - 2B instead.

Discussion Status

Some participants have offered hints regarding the distribution of A - 2B as a potential approach to the problem. There is an indication that the discussion is progressing towards a more manageable method, but no consensus or resolution has been reached yet.

Contextual Notes

The original poster expresses uncertainty about the next steps in the problem-solving process, particularly regarding the use of normal distribution properties and integration.

Gauss M.D.
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Homework Statement



The number of hours of sunshine in one week in a specific resort is assumed to follow a normal distribution with expectation 43 and standard deviation 17.

Family A will spend the first three weeks of the summer at the resort. Family B will spend the LAST two weeks of the summer at the resort.

Assuming everything is independent, what is the probability that family A will get atleast twice as much sunshine as family B?

Homework Equations



A = # of sunshine hours for family A
B = # of sunshine hours for family B

The Attempt at a Solution



E(A) = 3*E(X) = 129
σ(A) = sqrt(3)*σ(X) = 29.4

E(B) = 86
σ(B) = sqrt(2)*σ(X) = 24

So A ~ N(129, 29.4), B ~ N(86, 24)

If we let Z = A/B and try to find P(Z > 2), we do not have a linear combination of normally distributed random variables so I guess that's not the right way to go, assuming we want to do the normal table reading thing. So I don't know where to go from here. Help!
 
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You could try an integration over A or B, and see if that works.

Edit: Oh, I did not see this. Follow awkward's hint, that is way easier.
 
Last edited:
Hint: What can you say about the distribution of the random variable A - 2B?
 
awkward said:
Hint: What can you say about the distribution of the random variable A - 2B?

D'oh! Thank you so much!
 

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