Depending upon the climate, some can be lost to evaporation and plant uptake, but other things to consider are land use, soil type, and antecedent moisture condition.
For example, not much water will run off of dry sand. But water will just sheet off of a paved parking lot or saturated clayey loam.
I’m under the influence of cough syrup, but I can tell you that 99.99% of the precipitation falling on the lake will count as “runoff”. That is, if you have six inches of rain you will get six inches of runoff, on a lake’s surface.
Okay, lessee. (I’ve been doing this kind of stuff all week, so I’m on a roll…)
If we take the higher figure of 64 inches per year annual precipitation, that’s 1.63 metres (Wow. More than I thought.)
A lake surface area of 7340 square miles is 19.0 x 10[sup]9[/sup] m[sup]2[/sup].
So over the year, the lake directly receives a volume of 1.63 m * 19.0 x 10[sup]9[/sup] m[sup]2[/sup] = 31.0 x 10[sup]9[/sup] m[sup]3[/sup] of water.
But 31.0 x 10[sup]9[/sup] m[sup]3[/sup] per year = only 982 m[sup]3[/sup]/s if you average it out. Which, given the number of sunny days, means that the average rainstorm must put out a lot of water in a little time…
I assume that the whole lake surface gets 48 inches of rain a year. So, we’ll call it 1.22 meters. The area of the lake is 19,011 km[SUP]2[/SUP] or 19,011,000,000 m[SUP]2[/SUP]. So, that means the lake gets 23,178,000,000 m[SUP]3[/SUP] of rain a year, which works out to about 734 m[SUP]3[/SUP]/s or thereabouts. I don’t know how that compares to what you got, but I had fun calculating it, so there’s that.
The amount of rainfall varies over the length (time-wise) of the storm. This will vary depending upon where the rain storm is located (geographically). In the U.S., these distributions have been standardized for use in hydrologic computer modeling. I’m not familiar with Canadian hydrology but I assume something similar has been developed there.