Verification of linear combinations of a normal distribution A machined part consists of 5 independent components connected end-to-end. Two of these have lengths $N(37.0, 0.49)$, and three of these have lengths $N(24.0, 0.09)$. All measurements are in mm. What is the probability that the assembled part will be between 144 and 147 mm in length?
In answering this, I'm doing the following, but the answer is apparently not correct:
$$A = N(37.0, 0.49)$$
$$B = N(24.0, 0.09)$$
Linear combination...
$$Y = 2A + 3B$$
$$Y \sim N(2\cdot37.0 + 3\cdot24.0, 2^2\cdot0.49 + 3^2\cdot0.09)$$
$$Y \sim N(146.0, 2.77)$$
Therefore, the CDF between 144 and 147 should be:
$$\phi\left(\frac{147-146}{\sqrt{2.77}}\right)-\phi\left(\frac{144-146}{\sqrt{2.77}}\right) = 0.61128$$
The rational here is that the CDF at the specified points (144 and 147) gives $P(Y\le144)$ and $P(Y\le147)$, so we can subtract one from the other to get the cumulative probability between them. This makes intuitive and mathematical sense to me.
However, the answer given is 0.7777, and I'm not seeing how...

We have five independent random variables, $A_1,A_2,B_1,B_2,B_3$. The sum $Y$ of the lengths is given by $Y=A_1+A_2+B_1+B_2+B_3$. 
The mean of the sum is the sum of the means, and the variance of the sum is the sum of the variances. 
