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: $X\sim N(\mu, \sigma)$

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

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

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

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

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

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

$\mu=0.35+1.475791\sigma$

$\mu=0.6-1.226528\sigma$

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

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

$\mu\approx0.4865$

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