250 students complete a business college diploma. The results are normally distributed with a mean of 64% and a standard deviation of 8.6%. 15 students gain distinctions, 35 gain credits and 23 fail the course. Determine the cut-off percentages for each of the grades: a) Pass b) Credit c) Distinction

fail 0-x. mid values. =X/2

pass x-a. =(x+a)/2. pass no.=250-35-15-23=177

credit a-y. =(a+y)/2

distinction y-100. =(y+100)/2

mean=1/250*(x/2)×23+(x+a)/2*177+(y+a)/2×35+(y+100)/2×15=64

1/500[23x+177x+177a+35a+35y+15y+100*15)=64

200x+212a+50y+1500=32000

variance=[(X/2-64)^2*23+(x+a,/2-64)^2*277+(a+y/2-64)^2*35+(100+y/2-64)^2*15]/250=(8.6)^2

I've made two equations that has to be simplified.

probably a nother data has to be given since we have 3 variables to solve for(X,a,y)

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4