problem on probability calculation for sum of two probability distributions

Question

In a book I saw following problem. The answer at the back of the book is 0.75. But I am getting a different result. What am I doing wrong?

Problem: Let $X$ and $Y$ be independent Standard Uniform random variables. Find the probability of an event $\{0.5<(X+Y)<1.5 \}$.

My Solution:

Let $K=X+Y$.

$\text{E}(K)=0+0=0$.

$\text{Var}(K)=1+1=2$.

Let us standardize bounds:

$z_1 = (0.5-0)/\sqrt{2}=0.3536$.

$z_2=(1.5-0)/\sqrt{2}=1.0607$.

Then:

$P\{0.5<K<1.5\}=P\{0.3536<Z<1.0607\}=\Phi(1.0607)-\Phi(0.3536)=0.8584-0.6368=0.2216$

The $\Phi$ values are taken from a table in the back of the book.

Answer 1

The variables are uniform, not normal. You want the area in the unit square between the lines $x+y=0.5$ and $x+y=1.5$. That strip is all of the square except two right triangles of side $\frac12$, so its area is

$$1-2\cdot\frac18=\frac34\;.$$