Question

Reserve Problems Chapter 5 Section 6 Problem 1 Let X denote the active ingredient percentage (by weight) in a pharmaceutical capsule and suppose it is approximately normally distributed with a mean of 0.9% and a standard deviation of 0.02%. A new mixing process is Implemented and the mean percentage X of the active ingredient from a sample of 24 independent capsules is 0.91%. Determine P(x > 0.91) for the old mixing process. Round your answer to four decimal places (0.9. 9.8765). P(X 0.91) - the tolerance is +/-2% LINK TO TEXT that the new mixing process also has a mean active ingredient percentage at 0.9% and a standard deviation of 0.0247

Answer 1

Lel x be the achve ingredient percentage. in a phaimaceukical capsule ace XCNU:0.9%, 6: 0.02 .) Now, P z3) X~N, 6), then 6 O 02 0.0041 V24 Now, Plx91p- 1-0-3 O.O04 NPZ244 - o.00n3 from nmal prub. 1able . ghe el32031] fo dhe ald mixing pmce 0.00 13 is 2.44 Bioce the new mixing pncess has a meanis. 0.91 4 8tancland dleiahon is 6a0041% t is pot likely that new mixing pacey mean achve ingredient also eMea has q percentage is 0.g y. & sdandard alēviaho. O.O27.