Question

the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtain more than 75 marks and 20% of them obtained less than 40 marks. find the value of  μ  and σ​

Answer 1

Let
`X` 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

`P(X>75)=P\left(\frac{X-\mu}\sigma>\frac{75-\mu}\sigma\right)=P(Z>z_1)=0.10`

where
`Z` follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

`P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^(-1)(0.90)`

where
`F_Z(z)=P(Z\le z)` denotes the CDF of
`Z`, and
`F_Z^(-1)` denotes the inverse CDF. We have

`z_1=F_Z^(-1)(0.90)\approx1.2816`

Similarly, because 20% of students obtain less than 40 marks, we have

`P(X<40)=P\left(\frac{X-\mu}\sigma<\frac{40-\mu}\sigma\right)=P(Z<z_2)=0.20`

so that

`P(Z<z_2)=F_Z(z_2)=0.20\implies z_2=F_Z^(-1)(0.20)\approx-0.8416`

Then
`\mu,\sigma` are such that

`\frac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma`

`\frac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma`

and we find

`\mu\approx53.8739,\sigma\approx16.4848`