Question

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

362

375

389

415

358

409

379

380

367

366

(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	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 μ and σ? 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 ≤ 400) = Φ((400 − μ)/σ).]

(Round your answer to four decimal places.)

Answer 1

a) Since we know that

Where n is the number of data points
Now

and n = 10
This implies that

Since we know that

b) Mean () = 380
Sample size (n) = 10
Standard deviation (s) = 19.2815
Confidence interval(in %) = 95

Since we know that

Required confidence interval =
Required confidence interval = (380.0-13.78, 380.0+13.78)
Required confidence interval = (366.22, 393.78)

c)

This is a normal distribution question with

x = 400
P(x < 400.0)=?
The z-score at x = 400.0 is,

z = 1.0373
This implies that
P(x < 400.0) = P(z < 1.0373) = 0.8502
PS: you have to refer z score table to find the final probabilities.
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