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. a)Calculate the variance of X/50 1 b) Calculate the variance of 5X 3. 6250 c) Calculate the probability that the tensile strength of a tungsten component is at least 500 gscm? .8212 d) What is the probability that X is within 1 standard deviation of its mean? 6827 e) What is the 97.1st percentile of tensile strength of our tungsten component in gscm? 640.50 Suppose we make 3 independent strength measurements of tungsten components. What is the probability that all 3 measurements are at least 500? 9) 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

a)Variance of (X/50)=Var(X)/50²=50²/2500=1

b)Var(5X+3)=5²*Var(X)=25*50²=62500

c)P(X>500)=P(Z>(500-546)/50)=P(Z>-0.92)=0.8212

d)0.6827

f)P(all three are at least 500)=(0.8212)³ =0.5538

g)P(X>525|X>500)=P(X>525)/P(X>500)=P(Z>-0.42)/P(Z>-0.92)=0.6628/0.8212=0.8071

h)

varaince of Xbar =Var(X)/k

k=50²/0.8=3125