Question

The tensile strength, X, of a tungsten component is normally distributed with a mean of 546 grams per square centimeter (gscm) and a standard deviation of 50 gscm.

e) What is the 97.1st percentile of tensile strength of our tungsten component in gscm?

f) Suppose we make 3 independent strength measurements of tungsten components. What is the probability that all 3 measurements are at least 500?

g) What is the probability that X is greater than 525 given that X is greater than 500?

h) Suppose X1,X2,,Xk are k independent strength measurements of tungsten components. Let Xbar be the mean of those k values. How large must k be so the variance of the distribution of Xbar equals .8?

Answer 1

Solution:

We are given

Mean = µ = 546

SD = σ = 50

Part e

Z value for 97.1^st percentile = 1.895698

(by using z-table or excel)

X = Mean + Z*SD

X = 546 + 1.895698*50

X = 640.7849

Required answer = 640.7849

Part f

We have to find P(Xbar≥500) = 1 – P(Xbar<500)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (500 – 546)/[50/sqrt(3)]

Z = -46/28.86751

Z = -1.59349

P(Z<-1.59349) = P(Xbar<500) = 0.055526

P(Xbar≥500) = 1 – P(Xbar<500)

P(Xbar≥500) = 1 – 0.055526

P(Xbar≥500) = 0.944474

Required probability = 0.944474

Part g

We have to find P(X>525|X>500) = P(X>525) / P(X>500)

P(X>525) = 1 – P(X<525)

Z = (525 - 546)/50

Z = -0.42

P(Z<-0.42) = 0.337243

P(X>525) = 1 – P(X<525)

P(X>525) = 1 – 0.337243

P(X>525) = 0.662757

Now, we have

P(X>500) = 1 – P(X<500)

Z = (500 - 546)/50

Z = -0.92

P(Z<-0.92) = 0.178786

P(X>500) = 1 – P(X<500)

P(X>500) = 1 – 0.178786

P(X>500) = 0.821214

P(X>525|X>500) = P(X>525) / P(X>500)

P(X>525|X>500) = 0.662757/ 0.821214

P(X>525|X>500) = 0.807045

Required probability = 0.807045

Part h

We are given

Variance of the distribution of Xbar = .8

σ^2/k = 0.8

50^2/k = 0.8

2500/k = 0.8

k = 2500/0.8

k = 3125