Luminous Intensity, Iv for visible LED's is always peak maximum and then roughly 50% at 1/2 the BW angle to either side.
Your LED spec is 50° ±10° as the total beamwidth \$2θ^{1/2}\$ at half intensity
IR LED's often with very narrow θ were once all defined as \$θ^{1/2}\$ meaning the peak was half angle and not always dead centre. Recently to avoid newbie confusion, some IR specs show the full angle.
sage advice
A rule of thumb on diversity gain of the lens is that when you reduce the \$2θ^{1/2}\$ by 50% the Iv intensity doubles but due to lens loss -10% each time to magnify 2x or reduce the angle from no lens which is called the "Lambertian" response curve of 160° like most SMD LEDs.
Thus to compare your 50 ° LED should be about 50% of the Iv of an equivalent chip with 28~30 ° or other words a 30 ° Iv could be 2x {150~200mcd}[50deg] = 300~400 mcd
Now I only use 30° 5mm LEDs for most applications and only Iv > 10,000 mcd with tighter tolerances and get in bulk 10k MOQ but usually have lots left over in many colours and white > 16,000 mcd and unlike most LEDs these are Zener protected.
