Question

solve all parts please

The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.1 mm and standard deviation of 0.02 mm: (3 pts) (a) What is the probability that the diameter of the dot exceeds 0.066 mm? (b) What is the probability that the diameter is between 0.07 mm and 0.122 mm? (c) What standard deviation of diameters would be needed so that the probability in part (b) is 0.995?

Answer 1

1. Answer:. p ? t ' consider x: The diameter of the dot produced by pointer (923)...! given that x follows Normal distribution with it mean 40.1 mm Standard deviation sro 0.02 mm H P(x>0.066) pf X-ul > 0.066–4 L :: X-ell = 2.A N10,1) C PCzas 0.066 - 0.1 0.02 a Plz>=1:7) . : S I POZ -1.7) 1- Plz> 17) due to symmetry. 51 -0.0446 0.9554 Meronco, 1) Y from Normal probabilite / table. ZINITA probability that the diameter of the dot exceedis 0.066 mim is 0.9554

p(0.07 < X < -RC 0.07- 0.122) X-ell < 0.122 - 7- X- War NGO,L). - pl 0.07-01 L24 0.122-011_) 0.02. 0.02 = PO-1:5 < z < lal). - 1 P(Z <-1.5)=P(Z > 1.1) = 1- P(Z >1.5)- P(Z > 1:1) due to symmetry 1 0.0668 -0.1357 F 07975 DZONO,1) o from Normal probability MINN table -Co 15 to HINTAI probability that the diameter is between 0.07 mm and 0.122 mm is 0.7975 given that pl 0.07 < x < 0:122 ) - 0.995 if, all of ? p(0:07 < X< 0:122) = 0.995

0:07_0 Z < 0.122 - OL ogg5 -0.03 zے 0.022) -0.995 " r >0 -|-|( 2 > 0.03 ) -R42> 0.022 ) - 0.995 if r-0.0085 > N(0:1,57 then p10.07<x< 0.122) = 0.995 - 0.07 0.1 16:122 0.0085