f/2.2 is about 1/3 of a stop slower than an f/2.0 lens. The f stop is basically a ratio of the diameter of the aperture to the focal length and essentially gives you an idea of how much light is being let in to the sensor at a given aperture.
To make things a little simpler, f/2.8 is a full stop slower than f/2.0. That means that there is twice the amount of light that reaches the sensor at f/2.0 compared to f/2.8. To get the same exposure in the scene, you'll need to have a shutter speed that's twice as long, so in this case say 1/100 vs 1/200. With 1/3 of a stop it's not exactly a linear correlation because of how f-stops are calculated, but it works out that at 1/3 of a stop slower, you'll need about 80% of the equivalent shutter speed.
With the 1020 vs the 920, to get the same exposure, you'll need 1/100 shutter speed on the 920 vs 1/125 on the 1020.
But what if you want to use the same shutter speed and get the same exposure on both cameras? Well, since the aperture is fixed, on the 1020, you could increase the ISO sensitivity so that in effect it's 1/3 of a stop more sensitive to light. With that comes a tradeoff though - increased sensor noise.
For low light photography it may come more into play as the general rule of thumb for sharp photos is 1/effective focal length, so about 1/25 for both cameras. If you're on the cusp of that, it may mean the difference between a sharp photo and blurry photo (MAY mean the difference). Throw in another variable too - both cameras have image stabilization. This will allow you to hand hold the camera at slower shutter speeds and still attain a sharp photo. If the IS is better in the 1020 it could negate this factor, but also keep in mind that IS can only stop blur from camera shake, not motion of the subject.
In the end, what does all this really mean? IMO, in real world usage, 1/3 of a stop of difference is pretty negligible. I feel the advantages of the 1020 more than make up the difference (asa an owner of the 920 and 1020).