Note that the datasheet gives you the formula:

Nominal Transfer Value: VOUT = VS (P × 0.004 - 0.04)

If Vs = 5v, and you use the default reference voltage (5v) on the analog inputs, the Vout/Vs = (P x 0.004 - 0.04).

The value you get from analogRead is equal to (Vout/Vs)*1023. (Let's call this IN).

Now just do some algebra:

IN/1023 = P * 0.004 - 0.04;

P = (IN/1023 + 0.04)/0.004;

Cleaning this up a bit:

1/0.004 = 250;

250 * 0.04 = 10;

P = 250 * (IN/1023 + 0.04)

P = 250 * IN / 1023 + 10

Be sure in your code to force this to be floating-point math:

float P;

int IN;

IN = analogRead(pin);

P = 250 * IN / 1023[glow].0[/glow] + 10; // the .0 forces floating-point math instead of integer math.

BTW:

.00488 is 5/1024

This is a common mistake because indeed a 1-bit ADC has 1024 possible values. But keep in mind:

0V = 0

5V = 1023, NOT 1024.

Granted the error is small enough that most people wouldn't notice but it is additional error in the calculations.