Answer: 0.4551

## Explanation For a continuous probability density function $f(x)$, the probability that $x$ lies between $a$ and $b$ is given by: $$ P(a \leq x \leq b) = \int_a^b f(x) \, dx $$ Given: - $f(x) = \frac{x^2}{144000}$ for $0 \leq x \leq 200$ - We need $P(126 \leq x \leq 130)$ **Step-by-step calculation:** $$ \begin{aligned} P(126 \leq x \leq 130) &= \int_{126}^{130} \frac{x^2}{144000} \, dx \\ &= \frac{1}{144000} \int_{126}^{130} x^2 \, dx \\ &= \frac{1}{144000} \left[ \frac{x^3}{3} \right]_{126}^{130} \\ &= \frac{1}{144000} \left( \frac{130^3}{3} - \frac{126^3}{3} \right) \\ &= \frac{1}{144000} \cdot \frac{1}{3} \left( 130^3 - 126^3 \right) \\ &= \frac{1}{432000} \left( 2,197,000 - 2,000,376 \right) \\ &= \frac{1}{432000} \left( 196,624 \right) \\ &= 0.4551 \end{aligned} $$ **Verification of calculations:** - $130^3 = 2,197,000$ - $126^3 = 2,000,376$ - Difference = $196,624$ - $196,624 \div 432,000 = 0.4551$ Therefore, the probability that the bond price is between $126 and $130 is **0.4551**, which corresponds to option C.

Author: Nikitesh Somanthe