Let X be the random variable of IQ test. Then
\(\displaystyle{X}\sim{N}{\left({100},{15}^{{2}}\right)}\)

which means that \(\displaystyle\frac{{{X}-{100}}}{{15}}\sim{N}{\left({0},{1}\right)}\)

We need to find \(\displaystyle{P}{\left({X}\le{135}\right)}\)

We write this is a way to get the standard normal variable: \(\displaystyle{P}{\left({\left(\frac{{{X}-{100}}}{{15}}\right)}\le\frac{{{135}-{100}}}{{15}}\right)}={P}{\left({\left(\frac{{{X}-{100}}}{{15}}\right)}\le\frac{{7}}{{3}}\right)}\)

Let Ф denote the standard normal distribution, so \(\displaystyle{P}{\left({\left(\frac{{{X}-{100}}}{{15}}\right)}\le\frac{{7}}{{3}}\right)}=Ф{\left(\frac{{7}}{{3}}\right)}\simФ{\left({2.33}\right)}\sim{0.9901}\)

Thus, (c) is correct.

which means that \(\displaystyle\frac{{{X}-{100}}}{{15}}\sim{N}{\left({0},{1}\right)}\)

We need to find \(\displaystyle{P}{\left({X}\le{135}\right)}\)

We write this is a way to get the standard normal variable: \(\displaystyle{P}{\left({\left(\frac{{{X}-{100}}}{{15}}\right)}\le\frac{{{135}-{100}}}{{15}}\right)}={P}{\left({\left(\frac{{{X}-{100}}}{{15}}\right)}\le\frac{{7}}{{3}}\right)}\)

Let Ф denote the standard normal distribution, so \(\displaystyle{P}{\left({\left(\frac{{{X}-{100}}}{{15}}\right)}\le\frac{{7}}{{3}}\right)}=Ф{\left(\frac{{7}}{{3}}\right)}\simФ{\left({2.33}\right)}\sim{0.9901}\)

Thus, (c) is correct.