Find a real number c such that $𝑃(\bar X−\bar Y > 𝑐) = 0.3$

Question

Let $X_1,\ldots,X_5$ and $Y_1,\ldots,Y_8$ be two independent simple random samples such that $X_i\sim\mathcal N(m = 7,\sigma^2 = 50)$ and $Y_i\sim\mathcal N(m = 5,\sigma^2 = 24)$. Find a real number $c$ such that $𝑃(\bar X−\bar Y > c) = 0.3$

Should be an easy problem but I can’t understand if I’m solving it right. How to find this number c, that probability of $\bar X−\bar Y$ higher than c is equal 0.3? Please help

My attempt: Here I found variance of means and their difference: $Var(\bar X) = \delta^2/n = 50/5 = 10$;
$Var(\bar Y) = \delta^2/n = 24/8 = 3$; $\bar X \sim \mathcal N(7, 10)$;
$\bar Y \sim \mathcal N(5, 3)$; $\bar X−\bar Y \sim \mathcal N(2, 13)$; $(\bar X−\bar Y -2)/(√13) \sim \mathcal N(0, 1)$

Answer 1

All of your calculations so far are correct, to finish you can notice that for any real number $c$ : $$\begin{align}\mathbb P(\bar X - \bar Y > c) &= \mathbb P(\bar X - \bar Y - 2 > c - 2)\\ &=\mathbb P\left(\frac{\bar X - \bar Y - 2}{\sqrt{13}} > \frac{c-2}{\sqrt{13}}\right)\\ &=1-\Phi\left(\frac{c-2}{\sqrt{13}}\right) \end{align} $$ Where $\Phi$ is the Cumulative Distribution Function of the standard normal distribution.

Therefore $\mathbb P(\bar X - \bar Y > c) = 0.3$ iff $1-\Phi\left(\frac{c-2}{\sqrt{13}}\right) =0.3$. To solve your problem, you thus need to find $c$ such that $$1-\Phi\left(\frac{c-2}{\sqrt{13}}\right) =0.3 $$