Suppose X~POI(10), find $P[5I have worked out using the tables in the back of the book, but the back of the book says the answer is 3/5, whereas I got 0.8842. Any advice? 
 A: Just to be explicit, if $X \sim \mathrm{Poi}(\lambda)$, then 
$$\mathbf{P}\{a < X < b\} = \sum_{k = a + 1}^{b - 1} e^{-\lambda} \lambda^k/k!$$
Take $\lambda = 10, a = 5, b = 15$. Then this formula specializes to 
$$
\mathbf{P}\{5 < X < 15\} = \sum_{k=6}^{14} e^{-10} \frac{10^k}{k!} \approx 0.8494555641.
$$
You can get more digits with a calculator if you want. 
A: Suppose $X \sim \mathsf{Pois}(\lambda = 10),$ You seek
$P(5 < X < 15) = P(X \le 14) - P(X \le 5) =  0.8495,$ rounded to four places.
In R statistical software this can be computed as
ppois(14, 10) - ppois(5, 10)
[1] 0.8494556

Alternatively, $P(5 < X < 15) = \sum_{i=6}^{14} P(X = i).$
In R this is computed as follows:
sum(dpois(6:14, 10))
[1] 0.8494556

In the graph below you want the sum of the heights of the bars
between the vertical dotted lines:
plot(x, pdf, type="h", lwd=2, main="Bar plot of POIS(10)")
 abline(h=0, col="green2")
 abline(v=0, col="green2")
 abline(v = c(5.5,14.5), col="red", lwd=2, lty="dotted")


Before moving on to the next problem, please see how you can
get this answer from the tables in your book.
Addendum: You can get a reasonably good
approximate answer by using the normal approximation to a Poisson distribution. Standardize using $\mu = 10$ and $\sigma=\sqrt{10}$ and use continuity correction. Using standard normal tables, you should get close to the following result from R. (Such approximations are often
accurate to two places; better for larger $\lambda).$
diff(pnorm(c(5.5,14.5), 10, sqrt(10)))
[1] 0.8452711


plot(x, pdf, type="h", lwd=2, 
   main="Bar plot of POIS(10) with Normal Fit")
 abline(v=0, col="green2")
 abline(h=0, col="green2")
 abline(v = c(5.5,14.5), col="red", lwd=2, lty="dotted")
 curve(dnorm(x,10,sqrt(10)), add=T, col="blue", lwd=2)

