Q.The weight of randomly selected 500 adult persons from a region of a city follows normal distribution. If the average weight of these persons is 55 kg and its standard deviation is 7 kg then:

$\mu=55,\sigma=7$: (1) $P(-2<z<1)=0.8185 \Rightarrow 409$ persons; (2) $P(z<-2)=0.0228 \Rightarrow 11$ persons. OR: middle 60% $\Rightarrow \pm0.84\sigma$: ₹11,640–₹18,360.

Main part — 500 persons, $\mu = 55$ kg, $\sigma = 7$ kg.

(1) Weights between 41 kg and 62 kg. Standardise:

$z_1 = \frac{41-55}{7} = -2, \qquad z_2 = \frac{62-55}{7} = 1$

Area between $z=-2$ and $z=1$:

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

Number of persons:

$500 \times 0.8185 = 409.25 \approx 409 \text{ persons}$

(2) Weights less than 41 kg. Here $z = -2$:

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

Number of persons:

$500 \times 0.0228 = 11.4 \approx 11 \text{ persons}$

OR — 1000 employees, $\mu = 15000$, $\sigma = 4000$, middle 60% income range. …