Question

Question code 1A26 Suppose the diameter of a certain car component follows the normal distribution with a mean of 18mm and a standard deviation of 4mm. If we randomly select one of these components, find the probability of the following a) that its diameter is between 9mm and 27mm b) that its diameter is greater than 10mm

Answer 1

Let u be the diameter of a certain car component HNN cuo mean (u) 18 mm standard deviation co= 4mm as P [ 9 < x <27] For n=9 2 = n-M 9-18 o 4 E -2.25 2=-2.25 M = 18 2-2.25 z=0 For n=27 2=n-M = 27-18 r 4 2.25 Using a table area to the area to the left of z = -2,25 is 0.01222 and to the left of z=2.25 is 0.98778 .. Area Area of shaded region 0898778 -0.01222 = 0.97556 = 0.9756 iip[q<»<27] -0.9756 area . The probability that it's diameter is between 9mm and 29 mm is 0.9756

by P [2)10] for n=10 2=x-u T = 10 - 18 4 = -2 2=-2 M=18 2 0 0.0228 Using 2 table, area to the left of 2 = -2 is The entire area under the curve is 1 ... Area under the curve to the right of is 2 = -2 1 – 0.0228 - 0.9772 :. Area of shaded region is 0.9772 P[n 10] =0.9772 .:: The probability that it's than 10 mm is 10.9772 diameter is greater