Let X be random variable that represents salrie of some pediatric physician, \(\displaystyle{X}\sim{N}{\left(\mu,\sigma^{{2}}\right)}\), and with \(\displaystyle{Z}=\frac{{{X}-\mu}}{\sigma}\) standard normal we have:

\(\displaystyle{0.25}={P}{\left({X}{<}{180}\right)}={P}{\left({Z}{<}{\frac{{{180}-\mu}}{{\sigma}}}\right)}\Rightarrow{P}{\left({Z}{<}-{0.675}\right)}={0.25}\)

\(\displaystyle{0.25}={P}{\left({X}{>}{320}\right)}={P}{\left({Z}{<}{\frac{{{320}-\mu}}{{\sigma}}}\right)}\Rightarrow{P}{\left({Z}{>}{0.675}\right)}={0.25}\)

And from that:

\(\displaystyle{\frac{{{180}-\mu}}{{\sigma}}}=-{0.675}\Rightarrow{180}-\mu=-{0.675}\sigma\)

\(\displaystyle{\frac{{{320}-\mu}}{{\sigma}}}={0.675}\Rightarrow{320}-\mu={0.675}\sigma\)

If we sum both expression we get \(\displaystyle{180}-\mu+{320}-\mu={0}\Rightarrow{500}={2}{m}{y}\Rightarrow\mu={250}\) and given that

\(\displaystyle\sigma={\frac{{{70}}}{{{0.675}}}}={103.7}\)

a) \(\displaystyle{P}{\left({X}{<}{250}\right)}={P}{\left({Z}{<}{\frac{{{250}-{250}}}{{{103.7}}}}\right)}={P}{\left({Z}{<}{0}\right)}=\Phi{\left({0}\right)}={0.5}\)

b) \(\displaystyle{P}{\left({260}{<}{X}{<}{300}\right)}={P}{\left({\frac{{{260}-{250}}}{{{103.7}}}}{<}{Z}{<}{\frac{{{300}-{250}}}{{{103.7}}}}\right)}\)

\(\displaystyle={P}{\left({\frac{{{10}}}{{{103.7}}}}{<}{Z}{<}{\frac{{{50}}}{{{103.7}}}}\right)}\)

\(\displaystyle={P}{\left({0.096}{<}{Z}{<}{0.482}\right)}={P}{\left({Z}{<}{0.482}\right)}-{P}{\left({Z}{<}{0.096}\right)}\)

\(\displaystyle=\Phi{\left({0.482}\right)}-\Phi{\left({0.096}\right)}={0.15}\)