Let $D$ denote the diameter of a randomly selected tree. By assumption, $D\sim N(\mu,\sigma)$, where $\mu=8.8$, $\sigma=2.8$.
Let $Z=\frac{D-\mu}{\sigma}=\frac{D-8.8}{2.8}$. Then $D\sim N(0,1)$.
Note that $D=\sigma\cdot Z+\mu=2.8\cdot Z+8.8$
(a)
The probability that the diameter of a randomly selected tree will be at least $10$ (in.) is
\[
\begin{split}
P(D\geqslant 10)&=P(2.8\cdot Z+8.8\geqslant 10)=P(2.8\cdot Z\geqslant 1.2)\\
&=P(Z\geqslant 0.42857143)=1-P(Z < 0.42857143)=1-P(Z\leqslant 0.42857143)\\
&=1-\Phi(0.42857143)
\end{split}
\]
By Table A.3, we have $P(0.43)=0.6664$, hence
\[P(D\geqslant 10)=1-\Phi(0.42857143)\approx 1-0.6664=0.3336\]
Obviously, the probability that the diameter of a randomly selected tree will exceed $10$ (in.) is the same value, i.e., $0.3336$.
(b)
The probability that the diameter of a randomly selected tree will exceed $20$ (in.) is
\[
\begin{split}
P(D > 20)&=P(D\geqslant 20)=P(2.8\cdot Z+8.8\geqslant 20)=P(2.8\cdot Z\geqslant 11.2)\\
&=P(Z\geqslant 4)=1-P(Z < 4)=1-P(Z\leqslant 4)\\
&=1-\Phi(4)
\end{split}
\]
By Table A.3, we have $P(4)\approx 1$, hence
\[P(D\geqslant 20)=1-\Phi(4)\approx 1-1=0\]
That is, the probability that the diameter of a randomly selected tree will exceed $20$ (in.) is approximately $0$.
(c)
The probability that the diameter of a randomly selected tree will be between $5$ and $10$ (in.) is
\[
\begin{split}
P(5\leqslant D\leqslant 10)&=P(5\leqslant 2.8\cdot Z+8.8\leqslant 10)=P(-3.8\leqslant 2.8\cdot Z\leqslant 1.2)\\
&=P(-1.35714286\leqslant Z\leqslant 0.42857143)=P(0.42857143)-P(-1.35714286)\\
\end{split}
\]
By Table A.3, we have $P(0.43)=0.6664$ and $P(-1.36)=0.0869$, hence
\[P(5\leqslant D\leqslant 10)=P(0.42857143)-P(-1.35714286)\approx P(0.43)-P(-1.36)=0.6664-0.0869=0.5795.\]
That is, the probability that the diameter of a randomly selected tree will be between $5$ and $10$ (in.) is approximately $0.5795$.
(d)
If the interval $(8.8-c,8.8+c)$ includes $98\%$ of all diameter values, then
\[
P(8.8-c\leqslant D\leqslant 8.8+c)=98\%
\]
Note that $\mu=8.8$, and
\[
\begin{split}
8.8-c\leqslant D\leqslant 8.8+c &\Rightarrow -c\leqslant D-\mu\leqslant c\\
&\Rightarrow -\frac{c}{\sigma}\leqslant\frac{D-\mu}{\sigma}\leqslant\frac{c}{\sigma}\\
&\Rightarrow -\frac{c}{2.8}\leqslant Z\leqslant\frac{c}{2.8}
\end{split}
\]
Thus,
\[P(-\frac{c}{2.8}\leqslant Z\leqslant\frac{c}{2.8})=0.98\]
We have
\[
\begin{split}
0.98&=P(-\frac{c}{2.8}\leqslant Z\leqslant\frac{c}{2.8})=2\cdot P(0\leqslant Z\leqslant\frac{c}{2.8})\\
&=2\cdot(\Phi(\frac{c}{2.8})-\Phi(0))=2\cdot(\Phi(\frac{c}{2.8})-0.5)
\end{split}
\]
It infers that $\Phi(\frac{c}{2.8})=0.98/2+0.5=0.99$. By Table A.3, we have $\Phi(2.3+0.03)=0.9901$.
Hence, $\frac{c}{2.8}\approx 2.3+0.03=2.33$. Then, we have $c\approx 6.524$.
