Answer to Question #222211 in Statistics and Probability for wilson

Question

In an examination 7% of students score less than 35% marks and 89% of students score less than 60% marks. Find the mean and standard deviation, if the marks are normally distributed.

Answer 1

Let $X=$ student's mark: $X\sim N(\mu, \sigma)$

$$P(X<x_1)=P(Z<\dfrac{x_1-\mu}{\sigma})=0.07$$

$$P(X<x_2)=P(Z<\dfrac{x_2-\mu}{\sigma})=0.89$$

$$\dfrac{x_1-\mu}{\sigma}\approx-1.475791$$

$$\dfrac{x_2-\mu}{\sigma}\approx1.226528$$

$$x_1=0.35, \dfrac{0.35-\mu}{\sigma}\approx-1.475791$$

$$x_2=0.6, \dfrac{0.6-\mu}{\sigma}\approx1.226528$$

$$\mu=0.35+1.475791\sigma$$

$$\mu=0.6-1.226528\sigma$$

$$0.35+1.475791\sigma=0.6-1.226528\sigma$$

$$\sigma=\dfrac{0.6-0.35}{1.475791+1.226528}\approx0.0925$$

$$\mu\approx0.4865$$

$$\mu=48.65\%, \sigma=9.25\%$$