For the part B
we have the following information
![\begin{gathered} \operatorname{mean}=\bar{x}=45 \\ \text{standard deviation=}\sigma=3 \end{gathered}]()
To get the approximate percentage that remains in service between 48 and 51 months, we will use the formula

and then follow the steps below
Step 1.
We will find the z-score corresponding to both 48 and 51 months
For 48 months

for 51 months

The plot of the distribution is given below
From the plot above, we can observe that the probability is

In terms of percentage, the value is

Thus, we will have
