Question

The mean score of $1000$ students for an examination is $34$ and $S.D.$ 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 limit of the marks of the central $70$ $\%$ of the candidates:
$\left\{P\right[0<z<0.25]=0.0987P[0<z<1.63]=0.4484P[0<z<1.04]=0.35\}$

Answer 1

Given, $\mu =34,\sigma =16,N=1000$
(i) $P(30<X<60);Z=\frac{X-\mu }{\sigma }$
$\therefore X=30,Z_1=\frac{30-\mu }{\sigma }=\frac{30-34}{16}$
$=\frac{-4}{16}=0.25$
$Z_1=-0.25$
$Z_2=\frac{60-34}{16}=\frac{26}{16}=1.625$
$Z_2=1.63$ [Approximately]
$P(-0.25<Z<1.63)=P(0<Z<0.25)+P(0<Z<1.63)$ (due to symmetry)
$=0.0987+0.4484=0.5471$
$\therefore$ Number of students scoring between $30$ and $60$ $=0.5471\times 1000=547$
(ii) limit of central $70$ $\%$ of candidates:
Values of $Z_1$ from the area label for the area $0.35$$=-1.04$
[since $Z_1$ lies the left of $Z=0$]
Similarly $Z_2=1.04$
$\therefore Z_1=\frac{X-34}{16}=1.04Z_2=\frac{X-34}{16}=-1.04$
$X_1=16\times 1.04+34X_2=-1.04\times 16+34$
$=16.64+34X_2=-16.64+34$
$X_1=50.64X_2=17.36$
$\therefore 70$ $\%$ of the candidates score between $17.36$ and $50.64$.