Q.(A) This question is only for normal students.
The weight of randomly selected 500 adult persons from a region of a city follows normal distribution. The average weight of these persons is 55 kg and its standard deviation is 7 kg.

(A) $Z_{41}=-2,\ Z_{62}=1$: $P(-2<Z<1)=0.8185\Rightarrow\approx409$; $P(Z<-2)=0.0228\Rightarrow\approx11$. (B) Standard normal: mean 0, SD 1, symmetric bell curve, area 1.

(A) For normal students. $X\sim N(\mu=55,\ \sigma=7)$, $N=500$.

(i) Between 41 kg and 62 kg:

$Z_1=\frac{41-55}{7}=-2,\qquad Z_2=\frac{62-55}{7}=1.$

$P(-2<Z<1)=P(0<Z<2)+P(0<Z<1)=0.4772+0.3413=0.8185.$

Expected number $=500\times0.8185=409.25\approx\mathbf{409}$ persons.

(ii) Less than 41 kg: $Z=\dfrac{41-55}{7}=-2.$

$P(Z<-2)=0.5-P(0<Z<2)=0.5-0.4772=0.0228.$

Expected number $=500\times0.0228=11.4\approx\mathbf{11}$ persons.

(B) For blind students - properties of the standard normal distribution:

1. It is the distribution of $Z=\dfrac{X-\mu}{\sigma}$, with mean $0$ and standard deviation (and variance) $1$.
2. Its curve is bell-shaped and symmetric about $Z=0$; mean $=$ median $=$ mode $=0$.
3. The total area under the curve is $1$, with $0.5$ on each side of $0$. …