Answer to Question #222211 in Statistics and Probability for wilson

Question #222211

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.


1
Expert's answer
2021-08-03T15:18:40-0400

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


P(X<x1)=P(Z<x1μσ)=0.07P(X<x_1)=P(Z<\dfrac{x_1-\mu}{\sigma})=0.07

P(X<x2)=P(Z<x2μσ)=0.89P(X<x_2)=P(Z<\dfrac{x_2-\mu}{\sigma})=0.89

x1μσ1.475791\dfrac{x_1-\mu}{\sigma}\approx-1.475791

x2μσ1.226528\dfrac{x_2-\mu}{\sigma}\approx1.226528

x1=0.35,0.35μσ1.475791x_1=0.35, \dfrac{0.35-\mu}{\sigma}\approx-1.475791

x2=0.6,0.6μσ1.226528x_2=0.6, \dfrac{0.6-\mu}{\sigma}\approx1.226528

μ=0.35+1.475791σ\mu=0.35+1.475791\sigma

μ=0.61.226528σ\mu=0.6-1.226528\sigma

0.35+1.475791σ=0.61.226528σ0.35+1.475791\sigma=0.6-1.226528\sigma

σ=0.60.351.475791+1.2265280.0925\sigma=\dfrac{0.6-0.35}{1.475791+1.226528}\approx0.0925

μ0.4865\mu\approx0.4865

μ=48.65%,σ=9.25%\mu=48.65\%, \sigma=9.25\%


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