I have an example of a test where CD-DVD decides "Class 2" where the PIE Av is 5.37. The other values are within your limits.
Concerning the average 5 for a total of 60000, does not CD-DVD count by 8 blocks bunches ? If you accept this, then, I have read somewhere that a block has (excluding parity check data) 192 rows of 172 bytes, that is 33,000 bytes. Times 8, we have 264,000. Given that a DVD is about 4,700,000,000 bytes, it follows that in a DVD we have about 17,800 bunches of 8 blocks. With 5 errors per bunch the total is 17,800 x 5 = 89,000. It is not so far from your 60000 ? But maybe I am completely wrong about the block size