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