Question

(1 point) An automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed with a mean of 115 cm and a standard deviation of 4.9 cm. A. Find the probability that one selected subcomponent is longer than 117 cm. Probability = B. Find the probability that if 3 subcomponents are randomly selected, their mean length exceeds 117 cm. Probability = C. Find the probability that if 3 are randomly selected, all 3 have lengths that exceed 117 cm. Probability =

Answer 1

mean M = 115 o= = 4.9 Standard deviation X~ Norm (4,0) that Where Xv Random variable an important represents lengths of subcomponent random P(x>4+) Convert into normal Z Variable score = P(> X-M) 1. p(7 > 17-115 4,9 Plz 70.41) -P(770-41) Die 7-6-4) P(Z 20.41) tail left area of Z = score of 0.41 from Z-tables with Z row and of ou the Column of o.ol value CS Scanned with CamScanner 0-6591

P(x>110) n=3 2 o ku X-Norm (M/V) - P(2> = P(Z >0-707) Scamsetnam = 0 - 3409 P(x > 117) P(x > 117) P/ z> olon Z > 117 115 4.96 tall areas -1-left tall 7 score of 0.707 from tables Z- o z Z row of 0.01 middle Columns Il 0.7602 1(x>117) = 0.2398 Selected lengthe exceed CS Scanned with Plz 20.701) - of and the with os 0.0 and Л P(Z >0.70 Z=0-707 p(3 randomly

3, (0.2398) (0-7602) say P=0.2398 Let say this probability and con be solved binomial distribution by using p(x=r=n n r nor p q (randomly selected) where n=3 p=0-2398 0-7602 q=1P- 117 cm) P( 3 lengths exceed let say Rfta that variable X - Random of lengths represents number in 3 randomly selected exceed 3 3-3 P(x-3) Cz (in ney n' -() (0.2398790 (n-r)? Menal Probability = 0-0138 x=1 Scanned with CamScanner