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: premium, standard, and substandard. Light bulbs with a lifespan of 500 days or more are considered premium, those with a lifespan of less than 300 days are considered substandard, and the rest are considered standard.
Table 1
\begin{tabular}{|c|c|c|}
\hline Lifespan (days) & Frequency & Relative Frequency \\
\hline$[100,200)$ & 20 & 0.10 \\
\hline$[200,300)$ & 30 & $a$ \\
\hline$[300,400)$ & 70 & 0.35 \\
\hline$[400,500)$ & $b$ & 0.15 \\
\hline$[500,600)$ & 50 & 0.25 \\
\hline Total & 200 & 1 \\
\hline
\end{tabular}
(1) Based on the data in Table 1, write the values of $a$ and $b$;
(2) A person randomly purchases $n$ ($n \in \mathbf{Z}_{+}$) 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 purchases 3 light bulbs from this batch for use. If the above frequencies are used as probabilities, let $X$ represent the number of substandard 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 premium quality light bulbs, 100 standard quality light bulbs, and 50 defective light bulbs in the sample.
Thus, the ratio of premium quality, standard quality, and defective light bulbs is $50: 100: 50=1: 2: 1$.
Therefore, according to stratified sampling, the number of light bulbs purchased is $n=k+2k+k=4k\left(k \in \mathbf{Z}_{+}\right)$.
Thus, the minimum value of $n$ is 4.
(3) All possible values of $X$ are $0, 1, 2, 3$.
By the problem, the probability of purchasing one light bulb and it being defective is $0.1+0.15=0.25$.

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

Therefore, the distribution of the random variable $X$ is as shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|c|}
\hline$X$ & 0 & 1 & 2 & 3 \\
\hline$P$ & $\frac{27}{64}$ & $\frac{27}{64}$ & $\frac{9}{64}$ & $\frac{1}{64}$ \\
\hline
\end{tabular}

Thus, the expected value of $X$ is
$$
E(X)=0 \times \frac{27}{64}+1 \times \frac{27}{64}+2 \times \frac{9}{64}+3 \times \frac{1}{64}=\frac{3}{4} .
$$