applied maths (normal distribution)

a. Mean, μ = 0.6(80) + 0.4(72.5) = 77 ohms

Variance = 0.6(5)2 + 0.4(6)2 = 29.4

Standard deviation, σ = √29.4 = 5.4222

b. Let X be the random variable of the resistance of the items

P(defective items)

= P(X - μ < -2σ) + P(X - μ > 2σ)

= P(z < -2) + P(z > 2)

= 0.0228 X 2

= 0.0456

c. P(not more than two defective items)

= 1 - P(no defective item) - P(1 defective item) - P(2 defective items)

= 1 - (1 - 0.0456)10 - 10C1 (1 - 0.0456)9(0.0456) - 10C2 (1 - 0.0456)8(0.0456)2

= 0.008934

d. The range for normal batch

= (77 - 2(5.4222) , 77 + 2(5.4222))

= (66.1556 , 87.8444)

P(defective belongs to B)

= P[z < (66.1556 - 72.5) / 6] + P[z > (87.8444 - 72.5) / 6]

= P(z < -1.0574) + P(z > 2.5574)

= 0.1452 + 0.0053

= 0.1505

P(defective belongs to A)

= P[z < (66.1556 - 80) / 5] + P[z > (87.8444 - 80) / 5]

= P(z < -2.7689) + P(z > 1.5689)

= 0.0028 + 0.0583

= 0.0611

P(batch B│defective)

= P(defective belongs to B) / P(defective)

= 0.1505 / (0.1505 + 0.0611)

= 0.7112