Solution:

It is first of all evident that the two smallest sums, $32$ and $36$, are respectively $a+b$ and $a+c$. In the same way, the two largest sums, $48$ and $51$, are respectively $c+e$ and $d+e$. Thus $a+b=32$, $a+c=36$; subtracting the first equation from the second one obtains
$$c-b=4.$$
In the same way, from $d+e=51$, $c+e=48$ one obtains
$$d-c=3$$
and finally, subtracting (1) from (2), one obtains
$$d-b=7.$$
It remains to understand which sum corresponds to the value $37$. A priori, there are two possibilities: $37=a+d$ or $37=b+c$. If it were $37=a+d$, then by (3) one would have $37=a+d=(a+b)+(d-b)=32+7=39$, which is absurd. Therefore only the possibility $37=b+c$ remains. At this point it is easy to compute the values of $a, b$ and $c$. Indeed $2a=(a+b)+(a+c)-(b+c)=32+36-37=31$, from which $a=15.5$, $b=32-a=16.5$ and $c=37-b=20.5$. Once the value of $c$ is known, we can also compute $e=48-c=27.5$, from which $d=23.5$. With these values of $a, b, c, d, e$ it indeed turns out that the three smallest sums are $32, 36$ and $37$ and the two largest sums are $48$ and $51$, so the only possible value for $e$ is $27.5$.