Let $F$ denote the force. Then by assumption, $F\sim N(15,1.25)$. Where $\mu=15$, $\sigma=1.25$.
Let $Z:=\frac{F-\mu}{\sigma}=\frac{F-15}{1.25}$, then $Z\sim N(0,1)$.
Note that $F=\sigma\cdot Z+\mu=1.25\cdot Z+15$.
(a)
The probability that the force is at most $17$ kips is $P(F\leqslant 17)$.
\[
P(F\leqslant 17)=P(1.25\cdot Z+15\leqslant 17)=P(1.25\cdot Z\leqslant 2)=P(Z\leqslant 1.6)=\Phi(1.6)
\]
By Table A.3, we have $\Phi(1.6)=0.9452$. Thus, The probability that the force is at most $17$ kips is $0.9452$.
(b)
The probability that the force is between $10$ and $12$ kips is $P(10\leqslant F\leqslant 12)$.
\[
\begin{split}
P(10\leqslant F\leqslant 12)&=P(10\leqslant 1.25\cdot Z+15\leqslant 12)=P(-5\leqslant 1.25\cdot Z\leqslant -3)\\
&=P(-4\leqslant Z\leqslant -2.4)=\Phi(-2.4)-\Phi(-4)
\end{split}
\]
By Table A.3, we have $\Phi(-2.4)=0.0082$, $\Phi(-4)=0$. Thus, $P(10\leqslant F\leqslant 12)=0.0082-0=0.0082$.
That is, the probability that the force is between $10$ and $12$ kips is $0.0082$.
(c)
The probability that the force differs from $15.0$ kips by at most $2$ standard deviations is
\[
\begin{split}
P(|F-15|\leqslant 2\sigma)&=P(|F-\mu|\leqslant 2\sigma)=P(\frac{|F-\mu|}{\sigma}\leqslant 2)\\
&=P(|Z|\leqslant 2)=2\cdot P(0\leqslant Z\leqslant 2)=2\cdot(\Phi(2)-\Phi(0))\\
&=2\cdot(\Phi(2)-0.5)
\end{split}
\]
By Table A.3, we have $\Phi(2)=0.9772$. Hence,
\[
P(|F-15|\leqslant 2\sigma)=2\cdot(\Phi(2)-0.5)=2\cdot(0.9772-0.5)=0.9544.
\]
That is, the probability that the force differs from $15.0$ kips by at most $2$ standard deviations is $0.9544$.
