Let X be a random number selected from the interval. Then the probability density for the random variable X is
[tex]f_X(x)=\begin{cases}\dfrac1{10-6.35}=\dfrac1{3.65}\approx0.2740&\text{if }6.35\le x\le 10\\0&\text{otherwise}\end{cases}[/tex]
8 and 10 are the only even integers that fit the given criterion (6 is more than 0.25 away from 6.35), so that we're looking to compute
P(|X - 8| < 0.25) + P(|X - 10| < 0.25)
… = P(7.75 < X < 8.25) + P(9.75 < X < 10.25)
… = P(7.75 < X < 8.25) + P(9.75 < X < 10)
(since P(X > 10) = 0)
… = 0.2740 (8.25 - 7.75) + 0.2740 (10 - 9.75)
… = 0.2055