Question

The shear strength of each of ten test spot welds is determined, yielding the following data (psi).

381

358

367

409

375

415

362

361

367

389

(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.)

(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 μ and σ? Now use the invariance principle.] (Round your answer to two decimal places.)

(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 ≤ 400) = Φ((400 − μ)/σ).] (Round your answer to four decimal places.)

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

X	(X - X̄)²
381	6.760
358	416.160
367	129.960
409	936.360
375	11.560
415	1339.560
362	268.960
361	302.8
367	129.960
389	112.360

	X	(X - X̄)²
total sum	3784	3654.4
n	10	10

a)

average =    ΣX/n =    378.40(answer)

sample variance ,s²=    Σ(X - X̄)²/(n-1)=   406.04

using the method of maximum likelihood-

σ² = 1/n*(n-1)*s²   365.44
std dev ,σ=√σ²=   19.12(answer)

b)

µ =    378.4                  
σ =    19.11648503                  
proportion=   0.95                  
                      
Z value at    0.95   =   1.645   (excel formula =NORMSINV(α))      
z=(x-µ)/σ                      
so, X=zσ+µ=   1.645   *   19.11648503   +   378.4  
X   =   409.84(answer)

c)

µ =    378.4              
σ =    19.11648503              
left tailed                  
X ≤    400              
                  
Z =   (X - µ ) / σ =   1.13          
                  
P(X ≤   400   ) = P(Z ≤   1.13   ) =   0.8707(answer)

excel formula for probability from z score is =NORMSDIST(Z)