Q.The average monthly expense of students residing in university hostel is ₹2,000 and its standard deviation is ₹500. If the monthly expense of a student follows normal distribution then :

(i) $z = -2.5$ to $-1.5 \Rightarrow$ area $= 0.4938 - 0.4332 = 0.0606 = 6.06\%$. (ii) $z = -0.4 \Rightarrow$ area $= 0.5 + 0.1554 = 0.6554 = 65.54\%$.

GSEB Class-12 Statistics, Normal Distribution (area under the curve):

Given $\mu = 2000$, $\sigma = 500$. Standardize with $z = \dfrac{X - \mu}{\sigma}$.

(i) Between ₹750 and ₹1,250:

$z_1 = \frac{750 - 2000}{500} = \frac{-1250}{500} = -2.5$

$z_2 = \frac{1250 - 2000}{500} = \frac{-750}{500} = -1.5$

Required area $= P(-2.5 < z < -1.5) = P(1.5 < z < 2.5)$ (by symmetry):

$= \text{Area}(0 \text{ to } 2.5) - \text{Area}(0 \text{ to } 1.5) = 0.4938 - 0.4332 = 0.0606$

Percentage of students $= 0.0606 \times 100 = \mathbf{6.06\%}$.

(ii) More than ₹1,800:

$z = \frac{1800 - 2000}{500} = \frac{-200}{500} = -0.4$

Required area $= P(z > -0.4) = 0.5 + P(-0.4 < z < 0) = 0.5 + \text{Area}(0 \text{ to } 0.4)$:

$= 0.5 + 0.1554 = 0.6554$

Percentage of students $= 0.6554 \times 100 = \mathbf{65.54\%}$.

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