You are in a boat in a pool with a rock in your hand. You

depends on how big the pool is and how accurately you can measure the water rise

If the rock were neutrally buoyant the water level would remain the same. It is heavier than water which causes it to displace more than its own volume while in the boat compared to at the bottom of the lake. Therefore the water level of the lake would go down.

At the moment the rock leaves your hand the water falls in the pool. When the rock enters the water the water level of the pool rises again to the level it had before you threw it.

really? I thought the water would go up ?

These answers are troubling. The only correct answer so far is Ben. The water level goes down.

I'm not a mechanical engineer, but I did stay in a Holiday Inn Express last night. I hope no one else who has answered this question here (except for Homebrook) is a mechanical engineer.
When you add the volume of the rock and subtract the volume of water previously displaced by the boat+rock, there is no change in the water level of the pool.
Another way to answer is: The water level measured at the side of the pool remains the same. The boat becomes more buoyant and the water level measured at the side of boat falls.

The weight of the boat plus you plus the rock has already displaced the height of the water. The only time the water level will change will be when the rock is mid air.

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