Let [tex]X[/tex] be the random variable for the number of marks a given student receives on the exam.
10% of students obtain more than 75 marks, so
[tex]P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10[/tex]
where [tex]Z[/tex] follows a standard normal distribution. The critical value for an upper-tail probability of 10% is
[tex]P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)[/tex]
where [tex]F_Z(z)=P(Z\le z)[/tex] denotes the CDF of [tex]Z[/tex], and [tex]F_Z^{-1}[/tex] denotes the inverse CDF. We have
[tex]z_1=F_Z^{-1}(0.90)\approx1.2816[/tex]
Similarly, because 20% of students obtain less than 40 marks, we have
[tex]P(X<40)=P\left(\dfrac{X-\mu}\sigma<\dfrac{40-\mu}\sigma\right)=P(Z<z_2)=0.20[/tex]
so that
[tex]P(Z<z_2)=F_Z(z_2)=0.20\implies z_2=F_Z^{-1}(0.20)\approx-0.8416[/tex]
Then [tex]\mu,\sigma[/tex] are such that
[tex]\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma[/tex]
[tex]\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma[/tex]
and we find
[tex]\mu\approx53.8739,\sigma\approx16.4848[/tex]
