In a population of working couples,the man's weekly income X and the woman's weekly income Y are statistically independent and normally distributed.X has a mean of 250 and variance of 50,while Y has a mean of 200 and a variance of 60.What is the probability husband's income exceed his wife's income,on average?
what I tried to do was take
H0:u1−u2<=0
H1:u1−u2 >0
where u1:population mean income of men and u2:population mean income of women and under H0 compute,
And after that get get the p-value because p-value is the minimum value of significance level that would reject Ho.When H0 is rejected I get the value where man's income exceeds womens income.But how to get n1 and n2 values?And I don't know if I am going in the correct way.Any help to get this done please.
This is not a hypothesis testing problem, it is a pure probability problem. We want the probability that $X\gt Y$, so we want the probability that $W\gt 0$, where $W=X-Y$.
By our conditions, $W$ is normally distributed, with mean $250-200=50$, and variance $50+50=110$. We have $$\Pr(W\gt 0)=\Pr\left(Z\gt \frac{0-50}{\sqrt{110}}\right),$$ where $Z$ is standard normal.
By symmetry, this probability is the same as $\Pr \left(Z\lt \frac{50}{\sqrt{110} }\right)$. This probability will turn out to be close to $1$.