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sam_0017
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The Correlation.. !??
Bert and Ernie are trying to impress Lady Moneypenny with their surfing skills. Bert’s wave ride times are normally distributed with a mean of 15 seconds and a standard deviation of three seconds. Ernie spends more time drinking rather than practising his surfing, so his wave ride times are normally distributed with a mean 12 seconds and a standard deviation of 2 seconds. Due to natural changes in surfing conditions, Bert and Ernie’s wave ride times are positively correlated. Lady Moneypenny notices that Ernie’s wave ride time exceeds Bert’s 20% of the time. Determine the correlation between Bert and Ernie’s wave ride times.
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i tray to solve it be using this teaneck and i don't know is it correct :
since there are normally distributed :
B~(15,32) R~(12,22 )
and we have P(B>R)= 20%
lat X= B-R , X~(μ ,λ)
μ=E [X] = E-E[R] = 15-12 = 3
and also:
λ= Var[X]= Var[B-R] = Var+Var[R]-2cov(B,R).
=9+4-2cov(B,R) =13-2cov(B,R). ... (*)
and from :
P(B,R)=[itex]\frac{Cov(B,R)}{\sqrt{Var(B). Var(R)}}[/itex]
so:
Cov(B,R) = 6P(B,R)
by souping in (*):
λ=13-12P(B,R)
So now we have :
X~N(μ,λ) = X~N(3,13-12P(B,R))
and from the question P(B>R)= 20% so P(B-R>0)=20%
so we can say P(B-R>0)=P(X>0) = 0.2
... And here i stooped !?
can anyone help with this Question ??
Homework Statement
Bert and Ernie are trying to impress Lady Moneypenny with their surfing skills. Bert’s wave ride times are normally distributed with a mean of 15 seconds and a standard deviation of three seconds. Ernie spends more time drinking rather than practising his surfing, so his wave ride times are normally distributed with a mean 12 seconds and a standard deviation of 2 seconds. Due to natural changes in surfing conditions, Bert and Ernie’s wave ride times are positively correlated. Lady Moneypenny notices that Ernie’s wave ride time exceeds Bert’s 20% of the time. Determine the correlation between Bert and Ernie’s wave ride times.
---------
i tray to solve it be using this teaneck and i don't know is it correct :
since there are normally distributed :
B~(15,32) R~(12,22 )
and we have P(B>R)= 20%
lat X= B-R , X~(μ ,λ)
μ=E [X] = E-E[R] = 15-12 = 3
and also:
λ= Var[X]= Var[B-R] = Var+Var[R]-2cov(B,R).
=9+4-2cov(B,R) =13-2cov(B,R). ... (*)
and from :
P(B,R)=[itex]\frac{Cov(B,R)}{\sqrt{Var(B). Var(R)}}[/itex]
so:
Cov(B,R) = 6P(B,R)
by souping in (*):
λ=13-12P(B,R)
So now we have :
X~N(μ,λ) = X~N(3,13-12P(B,R))
and from the question P(B>R)= 20% so P(B-R>0)=20%
so we can say P(B-R>0)=P(X>0) = 0.2
... And here i stooped !?
can anyone help with this Question ??