Here's the question
The scores, $X_1$ and $X_2$, in papers $1$ and $2$ of an examination are normally distributed with means $24.3$ and $31.2$ respectively and standard deviations $3.5$ and $3.1$ respectively. The final mark for each candidate is found by calculating $2X_1 + 1.5X_2$. Find the probability that a random sample of 8 candidates will have a mean final mark of less than $60$.
This is what I have done so far:
Let $Y = 2X_1 + 1.5X_2$
$E(Y) = E(2X_1 + 1.5X_2)$
$E(Y) = 2E(X_1) + 1.5E(X_2)$
$E(Y) = 2\times24.3+1.5\times31.2$
$\therefore E(Y) = 95.4$
Then,
$Var(Y) = Var(2X_1 + 1.5X_2)$
$Var(Y) = 2^2Var(X_1) + 1.5^2E(X_2)$
$Var(Y) = 2^2\times3.5^2+1.5^2\times3.1^2$
$\therefore Var(Y) = 70.6225$
After that,
$\bar Y \sim N(95.4, \frac{70.6225}{8})$
$P(\bar Y < 60)$
$= P(Z < \frac{60 - 95.4}{\sqrt(\frac{70.6225}{8})})$
$= P(Z < -11.9145)$
$\approx 0$
However, in my textbook, the answer says 0.9351. So I'm not sure where my mistake occurred.
Can someone tell me what my mistake is? Thanks.