If $X_1$, $X_2$, and $X_3$ are independent normals with means $\mu_1,\mu_2,\mu_3$ and variances $\sigma_1^2$, $\sigma_2^2,\sigma_3^2$, then $X_1+X_2+X_3$ is normally distributed, mean $\mu_1+\mu_2+\mu_3$ and variance $\sigma_1^2+ \sigma_2^2+\sigma_3^2$.
In our case, the sum has mean $43$ and variance $9$, so standard deviation $3$. Now all that remains is an ordinary normal distribution calculation. Note that $40$ is $1$ standard deviation below the mean.
Remark: A generalization can be useful. Under the same conditions of normality and independence, $\sum_{i=1}^n a_iX_i$ has normal distribution, mean $\sum_{i=1}^n a_i\mu_i$, and variance $\sum_{i=1}^n a_i^2\sigma_i^2$.