Question #253083

William passed an aptitude test in order to be accepted in law at McGill university. his result was 450, but the mean is 400, and the standard deviation is 75. His friend cassandra passed a similar but not identical test, in law again, but at l'univeristé de Montreal. her mark was 300, the mean is 200, and standard deviation.

Only the top 15% are selected for the program, are they going to be selected?


Expert's answer

If a student gets a mark m, and the results have a normal distribution with mean value μ\mu, and standard deviation σ\sigma then the part of resautls which are lower than m is p=m12πσe(xμ)22σdx=mμσϕ(t)dt=Φ(mμσ)p = \int_{-\infty}^{m}\frac{1}{\sqrt{2 \pi} \sigma}e^{-\frac{(x-\mu)^2}{2 \sigma}} dx= \int_{-\infty}^{\frac{m-\mu}{\sigma}} \phi (t) dt = \Phi(\frac{m-\mu}{\sigma}) ,


where ϕ(t)\phi(t) is a standard normal distribution and Φ(t)\Phi(t) is its cumulative distribution function.

But Φ(mμσ)=12[1+erf(mμ2σ)]\Phi(\frac{m-\mu}{\sigma}) = \frac{1}{2}[1 + erf(\frac{m-\mu}{\sqrt{2}\sigma})]


Substitute values for William m = 450, μ=400\mu = 400 and σ=75\sigma = 75 obtain p=0.5[1+erf(50275)]=0.7475p = 0.5 [1 + erf(\frac{50}{\sqrt2 75})]=0.7475 , which is less than 75%, so William is not selected


Substitute values for Cassandra m = 300, μ=200\mu = 200 and σ=75\sigma = 75 obtain p=0.5[1+erf(100275)]=0.9088p = 0.5 [1 + erf(\frac{100}{\sqrt2 75})]=0.9088 , which is more than 85%, so Cassandra is selected


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