Question

(2) A process manufactures ball bearings with diameters that are normally distributed with mean 25.1 millimeters and standard deviation 0.08 millimeter. (a) To meet a certain specification, a ball bearing must have a diameter between 25.0 and 25.2 millimeters. Find the percentage of ball bearings that meet the specification. (b) Find the third quartile of the diameters.

Answer 1

Soluhion ! x! The diameter of the ball bearings follows normal distribution with mean - la 25.1 mm std deviation oa 0.08 mm The pdf of x is 1 ; -00<x<co feuss of incorp** (*** - <l <co oro a ball a diameter between 25.0 and 25.2 mm. must have Q.a) to meet certain specification, beanings Find the percentag bearings that meet the specification That is here we have to find P[25.0 < x < 25.2) of ball ? >> P[25.03X 325.2] - PIXS25.2] - PIXS25.0] by using excel, the probability calcuated as NORM. 1287 (25.2, 25.1, 0.08,TRUE()) NORM. D I ST ( 25.0, 25.1,0.08 , TRUE()) * 1 It gives converted in P[ 25.0 < x 5 25.2) - 0.7887 P[25.0 < x < 25.2] 78.87% percentage The porcentage of ball bearings that meet the specification is 78.87%

an @ b) Find third quartile of the diameters By det? of third quartile CP3) P[ X 3031 = 0.75 x : diameter Qg: Third quartile of X By using excel, the value of Pg is calcwared as NORM. I NY (0.75, 25.1, 0.08) corresponding problubiling of quartile It gives value Az 9 1 1 u = 25.153959 2 decimal Younded to = 25.15 mm 25.15 mm The third quartile of the diameter is :) Note: for Continuous dishi buhon any PLX <x] : PEX <x] ii) for normal dish bution, P[X3x] can be calculated = NORM. DIST (2, 4, 5, TRUE(1) m) for any probability P[ X 5a] = p , the value of a can be calculated as NORM. INV CP, ,)