Question

The shear strength of each of ten test spot welds is determined, yielding the following data (psi). 409 393 358 361 367 362 374 389 375 415 (a) Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood. (Round your answers to two decimal places.) average 380.3 psi standard deviation psi (b) Again assuming a normal distribution, estimate the strength value below which 95% of all welds will have their strengths. (Hint: What is the 95th percentile in terms of u and o? Now use the invariance principle.] (Round your answer to two decimal places.) psi (c) Suppose we decide to examine another test spot weld. Let X = shear strength of the weld. Use the given data to obtain the mle of P(X < 400). [Hint: P(X s 400) = 0((400 – u)/o).] (Round your answer to four decimal places.)

Answer 1

1)

sample variance s2 =					412.6778
point estimate of standard deviation σ =sqrt(n*s2/(n-1))=						19.27

b)

for 95th percentile , critical value of z =1.645

hence estimated strength value below which 95% of all welds will have their strengths=mean+z*standard deviation=

412.00

c)

P(X<400)=P(Z<(400-380.3)/19.272)=P(Z<1.02)=

0.8461

2)

c)

for 90th percentile , critical value of z =1.28

therefore value that separates the largest 10% of all values in the thickness distribution from the remaining 90% =mean+z*standard deviation =

1.7789

d)

P(X<1.6)=P(Z<(1.6-1.3463)/0.338)=P(Z<0.75)=

0.7734

e)

estimated standard error =s/√n =

0.0845