Explanation

To find the probability that the bond price is between $8.00 and $9.00, we need to integrate the probability density function (pdf) over the interval [8, 9].

Given: $f(x) = \frac{x}{8} - 0.75 \quad \text{for } 6 \leq x \leq 10$

Step 1: Set up the integral $P(8 \leq x \leq 9) = \int_{8}^{9} \left(\frac{x}{8} - 0.75\right) dx$

Step 2: Evaluate the integral $\int \left(\frac{x}{8} - 0.75\right) dx = \frac{x^2}{16} - 0.75x + C$

Step 3: Apply the limits $\left[\frac{x^2}{16} - 0.75x\right]_{8}^{9} = \left(\frac{9^2}{16} - 0.75 \times 9\right) - \left(\frac{8^2}{16} - 0.75 \times 8\right)$

Step 4: Calculate At x = 9: $\frac{81}{16} - 6.75 = 5.0625 - 6.75 = -1.6875$

At x = 8: $\frac{64}{16} - 6 = 4 - 6 = -2$

Step 5: Find the difference $(-1.6875) - (-2) = -1.6875 + 2 = 0.3125$

Step 6: Convert to percentage $`$0.31`25 \times 100% = 31.25%$$

Therefore, the probability that the bond price is between $8.00 and $9.00 is 31.250%, which corresponds to option C.

Verification: We can verify this is a valid probability by checking that the pdf integrates to 1 over the entire domain [6,10]: $\int_{6}^{10} \left(\frac{x}{8} - 0.75\right) dx = \left[\frac{x^2}{16} - 0.75x\right]_{6}^{10} = \left(\frac{100}{16} - 7.5\right) - \left(\frac{36}{16} - 4.5\right) = (6.25 - 7.5) - (2.25 - 4.5) = (-1.25) - (-2.25) = 1$