The shear strength of each of ten test spot welds is determined, yielding the fo

Question

The shear strength of each of ten test spot welds is determined, yielding the following data (psi). 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.) Again assuming a normal distribution, estimate the strength value below which 95% of all welds will have their strengths. [What is the 95th percentile in terms of mu and sigma? Now use the invariance principle.] (Round your answer to two decimal places.) 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

Explanation / Answer

b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.95      
          
Then, using table or technology,          
          
z =    1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    376.2      
z = the critical z score =    1.644853627      
s = standard deviation =    19.87      
          
Then          
          
x = critical value =    408.8832416   [ANSWER]

***********************

b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    400      
u = mean =    376.2      
          
s = standard deviation =    19.87      
          
Thus,          
          
z = (x - u) / s =    1.197785606      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   1.197785606   ) =    0.884499754 [ANSWER]
     

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.