9.+ 0/4 points |Previous Answers DevoreStat9 6.E.025 My Notes Ask Your Teacher T
ID: 3362743 • Letter: 9
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9.+ 0/4 points |Previous Answers DevoreStat9 6.E.025 My Notes Ask Your Teacher The shear strength of each of ten test spot welds is determined, yielding the following data (psi) 384 367 364 358 362 404 415 375 389 409 (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 standard deviation 96.69 765.4 X psi X 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 and ? 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: PX 400) = ((400-1)/o).] (Round your answer to four decimal places.) You may need to use the appropriate table in the Appendix of Tables to answer this question Need Help? Read It Talk to a Tutor Submit Answer Save Progress Practice Another VersionExplanation / Answer
9.
a.
Mean = Sum of observations/ Count of observations
Mean = (384 + 367 + 364 + 358 + 362 + 404 + 415 + 375 + 389 + 409 / 10) = 382.7
Variance
Step 1: Add them up
384 + 367 + 364 + 358 + 362 + 404 + 415 + 375 + 389 + 409 = 3827
Step 2: Square your answer
3827*3827 =14645929
…and divide by the number of items. We have 10 items , 14645929/10 = 1464592.9
Set this number aside for a moment.
Step 3: Take your set of original numbers from Step 1, and square them individually this time
384^2 + 367^2 + 364^2 + 358^2 + 362^2 + 404^2 + 415^2 + 375^2 + 389^2 + 409^2 = 1468517
Step 4: Subtract the amount in Step 2 from the amount in Step 3
1468517 - 1464592.9 = 3924.1
Step 5: Subtract 1 from the number of items in your data set, 10 - 1 = 9
Step 6: Divide the number in Step 4 by the number in Step 5. This gives you the variance
3924.1 / 9 = 436.0111
Step 7: Take the square root of your answer from Step 6. This gives you the standard deviation
20.8809
Answer:
mean (u) = 382.70
standard deviation (sd)= 20.88
b.
95% of all welds will have their strnghts (mean and standard deviation)
mean = 95% of u = 363.565 = 363.56
standard deviation = 95% of sd = 19.83
c.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 382.7
standard Deviation ( sd )= 20.88
P(X > 400) = (400-382.7)/20.88
= 17.3/20.88 = 0.8285
= P ( Z >0.8285) From Standard Normal Table
= 0.2037,
P(X < = 400) = (1 - P(X > 400)
= 1 - 0.2037 = 0.7963
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