How to calculate the value of the fixed resistor in a light dependent voltage divider?
JohnD is correct: the LDR and R1 need to be interchanged for the circuit to work as you want. Assuming you interchange the LDR and R1  (the LDR should be connected from the transistor base to emitter), here's a way to calculate the required value of R1: 

Let's say the the LED should turn off for a condition somewhere in between Case3 (LDR= 8.6kΩ) and Case4 (LDR= 2.5kΩ). Let's choose LDR=5kΩ as a suitable condition. For this condition, the base voltage of the transistor should be approximately at the turn-on Vbe (about .6V from the 2N3904 datasheet available here http://www.jameco.com/Jameco/Products/ProdDS/783421.pdf). So Vout= .6 = 5* LDR/(LDR+R1) = 5* 5k/(5k+R1). Solving for R1 gives R1~36kΩ.

Next we should check how much current can be supplied to the LED in full darkness:
In full darkness, the LDR resistance is very high, and approximately an open circuit. R1 will therefore supply the entire base current, calculated as Ib = (5-.6)/R1 = 122 µA. With a transistor gain of ~70, this would result in a maximum collector current through the LED of app 8.5 mA. This may be enough for a nightlight LED. If you find the LED is not bright enough, then using two transistors in a Darlington arrangement would provide a much higher current gain and thereby allow more LED current.