连续随机变量 $Y$ 服从分布 $\mathrm{N}(\mu, \sigma^2)$。已知 $\mathrm{P}(Y < -a) = \mathrm{P}(Y > 5a) = 0.1$。将 $\mu$ 和 $\sigma$ 分别表示为 $ka$ 和 $ma$ 的形式，其中 $k$ 和 $m$ 是待确定的常数。

1. Determine $\mu$ using the symmetry of the normal distribution.

Since $\mathrm{P}(Y < -a) = \mathrm{P}(Y > 5a)$ and the normal distribution is perfectly symmetrical, the mean $\mu$ must lie exactly halfway between $-a$ and $5a$.
$\mu = \frac{-a + 5a}{2} = 2a$

2. Standardise the variable $Y$ to form an equation involving $\sigma$.

$\mathrm{P}(Y < -a) = 0.1$
Using the standard normal variable $Z \sim \mathrm{N}(0, 1)$:
$\mathrm{P}\left(Z < \frac{-a - \mu}{\sigma}\right) = 0.1$
Substituting $\mu = 2a$:
$\mathrm{P}\left(Z < \frac{-a - 2a}{\sigma}\right) = 0.1$
$\mathrm{P}\left(Z < -\frac{3a}{\sigma}\right) = 0.1$

3. Use the inverse normal function to solve for $\sigma$.

Using a graphing calculator to find the inverse normal value corresponding to a cumulative probability of 0.1:
$-\frac{3a}{\sigma} \approx -1.28155$
$\sigma = \frac{-3a}{-1.28155}$
$\sigma \approx 2.34a \text{ (to 3 s.f.)}$