Mean of two normal distributed random variables does not add up [on hold]












-1














X is a normally distributed random variable with the expected value of 10 and Sd = 3.



Y is also a normally distributed random variable with the expected value of 20 and Sd = 4.



Question: P(X>Y)



I know that I should create a new variable D with SD = Sqrt(3^2+4^2) = 5
The problem is the mean, I believe it should be 10-20 = - 10



But the answer says the standardization should be:
1-Fz ((0 - (-6)) / 5) = 0.0228



Now how am I suppossed to get the mean of D to be -6?






statistics random-variables normal-distribution







share|cite|improve this question













put on hold as unclear what you're asking by Did, NCh, Lord Shark the Unknown, user91500, José Carlos Santos Jan 5 at 14:44


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.















  • If the mean of X is 10 and the mean of Y is 20, indeed the mean of X-Y is 10-20=-10. Which text says otherwise?
    – Did
    Jan 4 at 17:57










  • It's a pratice question for an exam, written by my professor. For the following question P(X>0.8Y) he uses the same answer only with another SD. So I don't think he did the same mistake twice hence I doubt myself
    – Lime3645
    Jan 4 at 18:06










  • For P(X>0.8Y), the relevant mean is the mean of X-0.8Y, which happens to be -6. Did you/they mix up the answers?
    – Did
    Jan 4 at 18:07










  • I guess it would be possible but for the following question P(X>0.8Y) Corr = 0.7 the answer is still the same, only difference is Sd.
    – Lime3645
    Jan 4 at 18:11
















-1














X is a normally distributed random variable with the expected value of 10 and Sd = 3.



Y is also a normally distributed random variable with the expected value of 20 and Sd = 4.



Question: P(X>Y)



I know that I should create a new variable D with SD = Sqrt(3^2+4^2) = 5
The problem is the mean, I believe it should be 10-20 = - 10



But the answer says the standardization should be:
1-Fz ((0 - (-6)) / 5) = 0.0228



Now how am I suppossed to get the mean of D to be -6?






statistics random-variables normal-distribution







share|cite|improve this question













put on hold as unclear what you're asking by Did, NCh, Lord Shark the Unknown, user91500, José Carlos Santos Jan 5 at 14:44


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.















  • If the mean of X is 10 and the mean of Y is 20, indeed the mean of X-Y is 10-20=-10. Which text says otherwise?
    – Did
    Jan 4 at 17:57










  • It's a pratice question for an exam, written by my professor. For the following question P(X>0.8Y) he uses the same answer only with another SD. So I don't think he did the same mistake twice hence I doubt myself
    – Lime3645
    Jan 4 at 18:06










  • For P(X>0.8Y), the relevant mean is the mean of X-0.8Y, which happens to be -6. Did you/they mix up the answers?
    – Did
    Jan 4 at 18:07










  • I guess it would be possible but for the following question P(X>0.8Y) Corr = 0.7 the answer is still the same, only difference is Sd.
    – Lime3645
    Jan 4 at 18:11














-1












-1








-1







X is a normally distributed random variable with the expected value of 10 and Sd = 3.



Y is also a normally distributed random variable with the expected value of 20 and Sd = 4.



Question: P(X>Y)



I know that I should create a new variable D with SD = Sqrt(3^2+4^2) = 5
The problem is the mean, I believe it should be 10-20 = - 10



But the answer says the standardization should be:
1-Fz ((0 - (-6)) / 5) = 0.0228



Now how am I suppossed to get the mean of D to be -6?






statistics random-variables normal-distribution







share|cite|improve this question













X is a normally distributed random variable with the expected value of 10 and Sd = 3.



Y is also a normally distributed random variable with the expected value of 20 and Sd = 4.



Question: P(X>Y)



I know that I should create a new variable D with SD = Sqrt(3^2+4^2) = 5
The problem is the mean, I believe it should be 10-20 = - 10



But the answer says the standardization should be:
1-Fz ((0 - (-6)) / 5) = 0.0228



Now how am I suppossed to get the mean of D to be -6?






statistics random-variables normal-distribution




statistics random-variables normal-distribution






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 17:38









Lime3645Lime3645

92




92




put on hold as unclear what you're asking by Did, NCh, Lord Shark the Unknown, user91500, José Carlos Santos Jan 5 at 14:44


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






put on hold as unclear what you're asking by Did, NCh, Lord Shark the Unknown, user91500, José Carlos Santos Jan 5 at 14:44


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • If the mean of X is 10 and the mean of Y is 20, indeed the mean of X-Y is 10-20=-10. Which text says otherwise?
    – Did
    Jan 4 at 17:57










  • It's a pratice question for an exam, written by my professor. For the following question P(X>0.8Y) he uses the same answer only with another SD. So I don't think he did the same mistake twice hence I doubt myself
    – Lime3645
    Jan 4 at 18:06










  • For P(X>0.8Y), the relevant mean is the mean of X-0.8Y, which happens to be -6. Did you/they mix up the answers?
    – Did
    Jan 4 at 18:07










  • I guess it would be possible but for the following question P(X>0.8Y) Corr = 0.7 the answer is still the same, only difference is Sd.
    – Lime3645
    Jan 4 at 18:11


















  • If the mean of X is 10 and the mean of Y is 20, indeed the mean of X-Y is 10-20=-10. Which text says otherwise?
    – Did
    Jan 4 at 17:57










  • It's a pratice question for an exam, written by my professor. For the following question P(X>0.8Y) he uses the same answer only with another SD. So I don't think he did the same mistake twice hence I doubt myself
    – Lime3645
    Jan 4 at 18:06










  • For P(X>0.8Y), the relevant mean is the mean of X-0.8Y, which happens to be -6. Did you/they mix up the answers?
    – Did
    Jan 4 at 18:07










  • I guess it would be possible but for the following question P(X>0.8Y) Corr = 0.7 the answer is still the same, only difference is Sd.
    – Lime3645
    Jan 4 at 18:11
















If the mean of X is 10 and the mean of Y is 20, indeed the mean of X-Y is 10-20=-10. Which text says otherwise?
– Did
Jan 4 at 17:57




If the mean of X is 10 and the mean of Y is 20, indeed the mean of X-Y is 10-20=-10. Which text says otherwise?
– Did
Jan 4 at 17:57












It's a pratice question for an exam, written by my professor. For the following question P(X>0.8Y) he uses the same answer only with another SD. So I don't think he did the same mistake twice hence I doubt myself
– Lime3645
Jan 4 at 18:06




It's a pratice question for an exam, written by my professor. For the following question P(X>0.8Y) he uses the same answer only with another SD. So I don't think he did the same mistake twice hence I doubt myself
– Lime3645
Jan 4 at 18:06












For P(X>0.8Y), the relevant mean is the mean of X-0.8Y, which happens to be -6. Did you/they mix up the answers?
– Did
Jan 4 at 18:07




For P(X>0.8Y), the relevant mean is the mean of X-0.8Y, which happens to be -6. Did you/they mix up the answers?
– Did
Jan 4 at 18:07












I guess it would be possible but for the following question P(X>0.8Y) Corr = 0.7 the answer is still the same, only difference is Sd.
– Lime3645
Jan 4 at 18:11




I guess it would be possible but for the following question P(X>0.8Y) Corr = 0.7 the answer is still the same, only difference is Sd.
– Lime3645
Jan 4 at 18:11










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