Probability for a sum of independent gamma random variables 
Suppose we are told that the weight of each gum ball (in centigrams) is given by the gamma distribution function, with $α=25$ and $β=2$.
  Find the probability that 100 gum balls will go over the limit of a standard package of $52g$.

I use central limit theorem to normalize to random variable Z, and I get the probability is around $0.42$, while my solution says $0.0228$.
 A: Let's try to avoid confusion about parameterizations, centigrams vs. grams, and normal approximation.
When using a gamma distribution, you need to start by being
clear about the parameterization. You seem to be using $\alpha$
for the shape parameter and $\beta$ for the scale parameter.
The rate parameter $\lambda$ is the reciprocal of the scale parameter:
$\lambda = 1/\beta.$
If $X_i$ are iid $\text{Gamma}(\text{shape} = \alpha = 25, \text{scale} = \beta = 2),$
then $T = \sum_{i=1}^{100} X_i \sim \text{Gamma}(\text{shape} = 2500, \text{scale} = 2),$
which can be seen by looking at moment generating functions.
Let's use centigrams throughout.
In R statistical software, it is possible to get exact probabilities
associated with a gamma distribution. You seek 
$$P(T > 5200)  = 1 - P(T \le 5200) = 0.023819.$$
This exact probability (not using a normal approximation) is found in R as follows:
 alpha = 25;  beta = 2
 1 - pgamma(5200, 100*alpha, 1/beta)
 ## 0.023819

Notice that the second gamma parameter in R is the rate.
So what I take to be the 'answerbook answer'  0.02275 is 
accurate to about three places. (I suppose it is found from a normal
approximation.) In the graph below, the area you seek is under
the gamma density curve to the right of the vertical line.

A: Assuming you are using the $k,\theta$ parameterization given on the Wikipedia page, then the mean would be $50$ centigrams and the variance would be $100$ centigrams$^2$. For $100$ samples the mean would be $50$ grams and the variance would be $1$ grams$^2$. $52$ grams would be $2$ standard deviations above the mean. This gives a probability of approximately $0.02275$ using the Normal distribution.
Your solution looks right. The answer you got would be right if the variance were $100$ grams$^2$, but $10000$ centigrams$^2=1$ grams$^2$.

After seeing BruceET's answer, I looked to see if Mathematica also had the ability to compute the Gamma distribution, and indeed it does.
N[1 - CDF[GammaDistribution[2500, 2], 5200]] gives $0.023819$
However, since the answer in the book matches the approximation using the Normal distribution, I assume they wanted it worked that way.
