Question

Precision manufacturing: A process manufactures ball bearings with diameters that are normally distributed with mean 25.2 millimeters and standard deviation 0.07 millimeter. Round the answers to at least four decimal places. (a) Find the 20 percentile of the diameters. (b) Find the 34th percentile of the diameters. (c) A hole is to be designed so that 1% of the ball bearings will fit through it. The bearings that fit through the hole will be melted down and remade. What should the diameter of the hole be? (d) Between what two values are the middle 90% of the diameters? Part: 0 / 4 Part 1 of 4 Find the 20 percentile of the diameters. The 20th percentile of the diameters is millimeters.

Answer 1

a)

We need to find 'a' such that P(X<a) = 0.2, where X~N(25.2, 0.07²)

From Standard normal table, we know that

P(Z < -0.842) = 0.2, So z_p = -0.842

P₂₀​​= ​μ+z_p​×σ

= 25.2+(-0.842×0.07)

= 25.141

b)

We need to find 'a' such that P(X<a) = 0.34, where X~N(25.2, 0.07²)

From Standard normal table, we know that

P(Z < -0.413) = 0.34, So z_p = -0.413

P₃₄​​= ​μ+z_p​×σ

= 25.2+(-0.413×0.07)

= 25.171

c)

We need to find 'a' such that P(X<a) = 0.01, where X~N(25.2, 0.07²)

From Standard normal table, we know that

P(Z < -2.326) = 0.01, So z_p = -2.326

P₁​= ​μ+z_p​×σ

= 25.2+(-2.326×0.07)

= 25.037

d)

We need to find 'a' and 'b' such that

P(X <a) = 0.05 and P(X <b) = 0.95

We know that P(Z < - 1.645 ) = 0.05

So, a ​= ​μ+z_p​×σ

= 25.2+(-1.645×0.07)

= 25.085

We know that P(Z < 1.645) = 0.95

So, b ​= ​μ+z_p​×σ

= 25.2+(1.645×0.07)

= 25.315

So, between 25.085 and 25.315, 90% of the diameters lie