Question

The shear strength of each of ten test spot welds is determined, yielding the following data (psi). 362 415 358 396 375 389 367 371 409 385 (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 382.70 psi standard deviation (b) Again assuming a normal distribution, estimate the strength value below which 95% of all welds will have their strengths. Mint: What is the 95th percentile in terms of and o? Now use the invariance principle] (Round your answer to two decimal places.) pa (c) Suppose we decide to examine another test spot weld. Let X-shear strength of the weld. Use the given data to obtain theme of PEX < 400). (Hint: PIX S400) = ((400 - ) 1 (Round your answer to four decimal places.)

Answer 1

X	(X - X̄)²
362	428.49
415	1043.29
358	610.09
396	176.89
375	59.29
389	39.69
367	246.490
371	136.890
409	691.690
385	5.290

	X	(X - X̄)²
total sum	3827	3438.10
n	10	10

a)

average=ΣX/n =    3827.000   /   10   =   382.7000
                  

std dev = √ [ Σ(X - X̄)²/(n)] =   √   (3438.1/10)   =  18.54

b)

Z value at    0.95   =   1.645   (excel formula =NORMSINV(   0.95   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.645   *   18.54 +   382.7  
X   =   413.20   (answer)          

c)

P( X ≤    400   ) = P( (X-µ)/σ ≤ (400-382.7) /18.54)      
=P(Z ≤   0.933   ) =   0.8246   (answer)