Question

The mean score of $1000$ students for an examination is $34$ and standard deviation is $16$.
(i) How many candidates can be expected to obtain marks between $30$ and $60$ assuming the normality of the distribution and
(ii) Determine the limits of the marks of the central $70\%$ of the candidates
$\left\{P\right[0<z<0.25]=0.0987;P[0<z<1.63]=0.4484;P[0<z<1.04]=0.35\}$

Answer 1

Mean $=\mu =34$ S.D $=\sigma =16$

$f\left(x\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{(x-\mu )}^2}{2{\sigma }^2}}$

$P(30\le x\le 60)=-f\left(60\right)+f\left(30\right)$

$=\frac{1}{\sqrt{2a\times 256}}[e^{-\frac{{(60-34)}^2}{2\times 256}}+e^{-\frac{{(30-34)}^2}{2\times 256}}]$

$\frac{1}{\sqrt{2a\times 256}}\left[0.702\right]=0.018$

$\therefore$ Number of students expected $=1000\times 0.018=18$