It's no longer continuous!

The random variable $Y$ has probability distribution $\mathbb{P}[Y = j] \; = \; \left\{ \begin{array}{lll} \tfrac{1}{A} & \qquad & 1 \le j \le A-1 \\ \tfrac{1}{2A} & & j = 0,A \\ 0 & & \mbox{otherwise} \end{array} \right.$ and so $\begin{aligned} \mathbb{E}[Y] & = \tfrac{1}{A}\sum_{j=1}^{A-1}j + \tfrac{1}{2A} \times A \; = \; \tfrac12A \\ \mathbb{E}[Y^2] & = \tfrac{1}{A}\sum_{j=1}^{A-1}j^2 + \tfrac{1}{2A} \times A^2 \; = \; \tfrac16(2A^2 + 1) \\ \mathrm{Var}[Y] & = \tfrac{1}{12}(A^2 + 2) \end{aligned}$ Since this variance is $225.5$ , we deduce that $A = \boxed{52}$ .

The distribution of probability for $Y$ will have a probability $P$ for values $0$ and $A$ and $2P$ for all values $1,2,3,...A-2,A-1$ . This is because for a value of $X$ to be rounded as $0$ or $A$ , it has to be in $[0,0.5)$ and $[A-0.5,A]$ , respectively. For all ohter values $Y$ , the number can be in $[Y-0.5,Y+0.5)$ , which is twice as big.

So, first to determine the value of $P$ . All the probabilities must sum up to $1$ . The value is $P$ for $2$ of the possible values and $2P$ for the other $A-1$ possible values:

$\large \displaystyle P\cdot2 + 2P\cdot (A-1) = 1$

$\color{#20A900} \boxed{\large \displaystyle P = \frac{1}{2A}}$

So:

$\large \displaystyle p_i = \begin{cases} \frac{1}{2A}, i = 0, A \\ \frac1A, i = 1, 2, ... , A-1 \end{cases}$

The expected value $\mu$ is:

$\large \displaystyle \mu = \sum_{i=0}^A p_i \cdot y_i$

$\large \displaystyle \mu = \frac{1}{2A}\cdot (0 + A) + \sum_{i=1}^{A-1} \frac1A \cdot i$

$\large \displaystyle \mu = \frac12 + \frac1A \cdot \frac{A(A-1)}{2}$

$\color{#20A900} \boxed{\large \displaystyle \mu = \frac{A}{2}}$

Last, the variance:

$\large \displaystyle \sigma^2 = \sum_{i=0}^A p_i \cdot y_i^2 - \mu^2$

$\large \displaystyle \sigma^2 = \frac{1}{2A} \cdot (0^2 + A^2) + \sum_{i=1}^{A-1} \frac{1}{A} \cdot i^2 - \frac{A^2}{4}$

$\large \displaystyle \sigma^2 =\frac{A}{2} + \frac1A \cdot \frac{(A-1)\cdot A \cdot (2A-1)}{6} - \frac{A^2}{4}$

$\color{#20A900} \boxed{ \large \displaystyle \sigma^2 = \frac{A^2 + 2}{12} }$

Since the variance is equal to $225.5$ , then:

$\large \displaystyle \frac{A^2+2}{12} = 225.5$

$\large \displaystyle A^2 + 2 = 2706$

$\large \displaystyle A^2 = 2704$

Since $A$ is positive:

$\color{#3D99F6} \boxed{ \large \displaystyle A = 52}$