There are 26 integers in the set \(\{630,631,\ldots,655\}\). If the wavelength of light is uniformly distributed, then the probability that it has a wavelength of any one of these integers is \(\dfrac1{26}\), so the distribution of wavelengths has probability mass function
\(\mathbb P(X=x)=\begin{cases}\dfrac1{26}&\text{for }630\le x\le655,x\in\mathbb Z\\\\0&\text{otherwise}\end{cases}\)
The mean (expected value) is given by
\(\mathbb E(X)=\displaystyle\sum_xx\mathbb P(X=x)\)
and the variance by
\(\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2\)
First, the mean:
\(\mathbb E(X)=\displaystyle\sum_{x=630}^{655}x\mathbb P(X=x)=\frac1{26}\sum_{x=630}^{655}x\)
\(\mathbb E(X)=\dfrac{16705}{26}\approx643\)
(rounded up from an exact value of 642.5)
Now for the variance:
\(\mathbb E(X^2)=\displaystyle\sum_xx^2\mathbb P(X=x)=\frac1{26}\sum_{x=630}^{655}x^2\)
\(\mathbb E(X^2)=412862.5\)
Meanwhile, \(\mathbb E(X)^2=642.5^2=412806\), which gives a total variance of
\(\mathbb V(X)=412862.5-412806\approx57\)
(rounded up from 56.5)
