We are given two pieces of intormation:
1. The value 120 is at the 10 th percentile. This means $\(P(X \leq 120)=0.10\)$.
2. The value 280 is at the 40 th percentile. This means $\(P(X \leq 280)=0.40\)$.

We need to find the value at the 25th percentile.
For a standard normal distribution $\(Z \sim N(0,1)\)$, we can find the z-scores corresponding to these percentiles.
Let $Z_p$ be the z-score for the $p$-th percentile.
From a Z-table or calculator:
- For the 10th percentile (0.10 probability): $\(Z_{0.10} \approx-1.28\)$
- For the $\(\mathbf{4 0}\)$ th percentile ( $\(\mathbf{0 . 4 0}\)$ probability): $\(Z_{0.40} \approx-0.25\)$

We can set up two equations using the formula $\(X=\mu+Z \sigma\)$ :
1. $\(120=\mu+(-1.28) \sigma \Longrightarrow 120=\mu-1.28 \sigma\)$
2. $\(280=\mu+(-0.25) \sigma \Longrightarrow 280=\mu-0.25 \sigma\)$

Now we have a system of two linear equations with two variables ( $\(\mu\)$ and $\(\sigma\)$ ).
Subtract equation (1) from equation (2):

$$
\(\begin{aligned}
& (280-120)=(\mu-0.25 \sigma)-(\mu-1.28 \sigma) \\
& 160=-0.25 \sigma+1.28 \sigma \\
& 160=1.03 \sigma \\
& \sigma=\frac{160}{1.03} \approx 155.34
\end{aligned}\)
$$


Substitute $\(\sigma\)$ back into equation (1) to find $\(\mu\)$ :

$$
\(\begin{aligned}
& 120=\mu-1.28(155.34) \\
& 120=\mu-198.8352 \\
& \mu=120+198.8352=318.8352
\end{aligned}\)
$$

Now, we need to find the value at the 25 th percentile.
For the 25th percentile ( 0.25 probability): $\(Z_{0.25} \approx-0.67\)$ (or -0.6745 for more precision).
Let $\(X_{0.25}\)$ be the value at the 25 th percentile.

$$
\(\begin{aligned}
& X_{0.25}=\mu+Z_{0.25} \sigma \\
& X_{0.25}=318.8352+(-0.67)(155.34) \\
& X_{0.25}=318.8352-104.0778 \\
& X_{0.25}=214.7574
\end{aligned}\)
$$


So, Quantity A is approximately 214.76.
Quantity B is 200.
Comparing the two quantities:

$$
\(214.76>200\) .
$$


Therefore, Quantity A is greater than Quantity B.
Important Note on Z-scores and precision:
The exact Z-scores are crucial for normal distribution problems. Using more precise Z-scores:

$$
\(\begin{aligned}
& Z_{0.10} \approx-1.28155 \\
& Z_{0.40} \approx-0.25335 \\
& Z_{0.25} \approx-0.67449
\end{aligned}\)
$$


Let's recalculate with higher precision:
1. $\(120=\mu-1.28155 \sigma\)$
2. $\(280=\mu-0.25335 \sigma\)$

Subtracting (1) from (2):

$$
\(\begin{aligned}
& 160=(-0.25335-(-1.28155)) \sigma \\
& 160=(1.0282) \sigma \\
& \sigma=\frac{160}{1.0282} \approx 155.6117
\end{aligned}\)
$$


Substitute $\(\sigma\)$ into (1):

$$
\(\begin{aligned}
& 120=\mu-1.28155(155.6117) \\
& 120=\mu-199.426 \\
& \mu=120+199.426=319.426
\end{aligned}\)
$$

Now, calculate $\(X_{0.25}\)$ :

$$
\(\begin{aligned}
& X_{0.25}=319.426+(-0.67449)(155.6117) \\
& X_{0.25}=319.426-105.066 \\
& X_{0.25}=214.36
\end{aligned}\)
$$


Even with higher precision, Quantity A is approximately 214.36 , which is still greater than Quantity B (200).

The final answer is Quantity A is greater.