Question

The mean weight of $500$ male students in a certain college is $151$ pounds and the standard deviation is $15$ pounds. Assuming the weights are normally distributed, find the approximate number of students weighing.
(i) between $120$ and $155$ pounds,

$Z$	$0.2667$	$2.067$	$2.2667$
Area	$0.1026$	$0.4803$	$0.4881$

(ii) more than $185$ pounds.

Answer 1

Here $\mu =51$ and $\sigma =15$
$Z=\frac{X-\mu }{\sigma }=\frac{X-151}{15}$
When $X=120$, $Z=\frac{120-151}{15}=-2.067$
When $X=155$, $Z=\frac{155-151}{15}=0.2667$
When $X=185$, $Z=\frac{185-151}{15}=\frac{35}{15}=2.2667$
(i) $P(120<X<155)=P(-2.07<Z<0.27)$
$={\int }_{-2.067}^{0.2667}\varphi \left(z\right)dz={\int }_0^{0.2667}\varphi \left(z\right)dz+{\int }_0^{2.0667}\varphi \left(z\right)dz$
$=0.4803+0.1026=0.5829$.
Here $N=500$
$\therefore$ The number of students whose weight is between $120$ and $155$ pounds is
$0.5872\times 500=291$
(ii) $P(X<185)=P(Z>2.27)={\int }_{2.2667}^{\infty }\varphi \left(z\right)dz$
$={\int }_0^{\infty }\varphi \left(z\right)dz-{\int }_0^{2.2667}\varphi \left(z\right)dz=0.5-0.4881=0.0119$
$\therefore$ the number of students with weight about $185$ pounds.
$=0.0119\times 500=$ (approx) $6$.