19. (12 points) In a certain batch of a certain type of light bulb, 200 samples were randomly selected, and their lifespans were tracked and recorded in a frequency distribution table as shown in Table 1. Based on their lifespan, the light bulbs are classified into three grades: excellent, standard, and defective. Light bulbs with a lifespan of 500 days or more are classified as excellent, those with a lifespan of less than 300 days are classified as defective, and the rest are classified as standard.
(1) Based on the data in Table 1, write the values of $a$ and $b$;
(2) A person randomly purchased $n$ $\left(n \in \mathbf{Z}_{+}\right)$ light bulbs from the sample. If the grade distribution of these $n$ light bulbs is the same as that obtained by stratified sampling according to the three grades, find the minimum value of $n$;
(3) A person randomly purchased 3 light bulbs from this batch. If the above frequencies are used as probabilities, and $X$ represents the number of defective light bulbs purchased by this person, find the distribution of $X$ and its expected value.
19. (1) $a=0.15, b=30$.
(2) From Table 1, we know that there are 50 excellent lamps, 100 good lamps, and 50 defective lamps in the sample of lamps.
Thus, the ratio of excellent, good, and defective lamps is $50: 100: 50=1: 2: 1$.
Therefore, according to stratified sampling, the number of lamps purchased is
$$
n=k+2k+k=4k\left(k \in \mathbf{Z}_{+}\right) \text {. }
$$

Thus, the minimum value of $n$ is 4.
(3) All possible values of $X$ are $0,1,2,3$.

From the problem, we know that the probability of purchasing one lamp and it being defective is $0.1+0.15=0.25$.

Purchasing 3 lamps from this batch can be considered as 3 independent repeated trials, thus,
$$
\begin{array}{l}
P(X=0)=\mathrm{C}_{3}^{0}\left(1-\frac{1}{4}\right)^{3}=\frac{27}{64}, \\
P(X=1)=\mathrm{C}_{3}^{1} \frac{1}{4}\left(1-\frac{1}{4}\right)^{2}=\frac{27}{64}, \\
P(X=2)=\mathrm{C}_{3}^{2}\left(\frac{1}{4}\right)^{2}\left(1-\frac{1}{4}\right)=\frac{9}{64}, \\
P(X=3)=\mathrm{C}_{3}^{3}\left(\frac{1}{4}\right)^{3}=\frac{1}{64} .
\end{array}
$$

Therefore, the distribution of the random variable $X$ is as shown in Table 2. Thus, the expected value of $X$ is
$$
E(X)=0 \times \frac{27}{64}+1 \times \frac{27}{64}+2 \times \frac{9}{64}+3 \times \frac{1}{64}=\frac{3}{4} .
$$