If the grades are normally distributed with mean % of marks 75 and S. D. 10, determine the lowest % of marks to receive grade A and the lowest % of marks that passes.

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 08

Year : MAY 2014

Mean $(m)=75,$ Standard derivation $\sigma =10$

Let x denote marks obtained by a student on 100.

ii) Let x, be the lowest marks of the top $15\%$ (Grade A) Let $z_1$ be corresponding SNV

proportion of the top $15\%=0.15$

$\therefore $ Area between $‘z=0’$ and $‘z=z1’=0.5-0.15=0.35 $

$\therefore $ from z-table, $z_1=1.0364$

$\therefore \dfrac {x_1-m}6=1.0364\\ \therefore \dfrac {x_1-75}{10}=1.0364\\ \therefore x_1=85.3643\\ \therefore x_1=85$

Hence, the lowest marks to receive grade $A=85\%.$

ii) Let $x_2$ be the heist marks of the bottom $20\%$ or the lowest % of marks that passes.

Let $z_2$ be corresponding SNV.

Proportion of the bottom $20\%=0.20$

$\therefore $ Area between $‘z=0’$ to $‘z=z2’=0.5-0.2=0.3$

$\therefore $ From z table, $z_2=-0.8416$

$\therefore \dfrac {x_2-m}6=-0.8416\\ \therefore \dfrac {x_2-75}{10}=-0.8416\\ \therefore x_2=75+10(-0.8416)\\ \therefore x_2=62.5838\\ \therefore x_2=63$

Hence the lowest% of marks that passes $=67\%$

Ans: Lowest, arks to receive grade A $=85\%$

Lowest % of marks that passes $=67\%.$