A turbine shaft is made up of $4$ different sections. The lengths of those sections are independent and have normal distributions with $\mu$ and $\sigma$: (8.10, 0.22), (7.25, 0.20), (9.75, 0.24), and (3.10, 0.20). What is the probability an assembled shaft meets the speciications $28\pm 0.26$?
When we assemble the shaft we get a normal distribution with $\mu=28.2$ and $\sigma=0.43$. Call this random variable $Y$. Now we are trying to find$$P(27.74\leq Y\leq 28.26)$$which is $$P(-1.069\leq Z\leq 0.1395)=0.3995$$
but the textbook gives $0.4314$. This seems too significant to be a roundoff error in the normal distribution table.