Solution. The rounding error of the measurement can be considered as a random variable $X$, which is uniformly distributed in the interval between two adjacent integer divisions. The density of the uniform distribution $f(x)=1 /(b-a)$, where $(b-a)$ is the length of the interval in which the possible values of $X$ are contained; outside this interval, $f(x)=0$. In the problem under consideration, the length of the interval in which the possible values of $X$ are contained is 0.1, so $f(x)=1 / 0.1=10$. It is easy to understand that the measurement error will exceed 0.02 if it is contained in the interval $(0.02, 0.08)$.

Using the formula $P(a<X<b)=\int_{a} f(x) \mathrm{d} x$, we get

$$
P(0.02<X<0.08)=\int_{0.02}^{0.08} 10 \mathrm{~d} x=0.6
$$
308. The scale division value of the amperemeter is 0.1 A. The readings of the amperemeter are rounded to the nearest whole division. Find the probability that an error exceeding $0.02 \mathrm{~A}$ will be made during the reading.