The shear strength of each of ten test spot welds is determined, yielding the following data (psi). 409 378 358 362 389 404 415 375 367 396 (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.)

a. Average shear strength = 385.3 ; standard deviation of shear strength is ±19.25

b. At 95%, X = 1.645σ + μ

c. P(X ≤ 400) = 27.77% = 0.2777

Step-by-step explanation:

Mean = (409 + 378 + 358 + 362 + 389 + 404 + 415 + 375 + 367 + 396)/10

Mean = 3853/10

Mean = 385.3

Therefore, Average shear strength = 385.3

Variance = (409 - 385.3)² + (378  - 385.3)²+ (358  - 385.3)² + (362  - 385.3)²+ (389  - 385.3)²+ (404  - 385.3)²+ (415  - 385.3)²+ (375  - 385.3)²+ (367  - 385.3)²+ (396 - 385.3)²/10

Variance = 3704.1/10 = 370.41

Standard deviation = √370.41 = 19.25

Therefore, standard deviation of shear strength is ±19.25

b. Using normal distribution table at 95%, z = 1.645

z= (X - μ)/σ  

Therefore X = 1.645σ + μ

c. X = 400; μ= 385.3 ; σ = 19.25

z= (X - μ)/σ

z= (400 - 385.3)/19.25

z = 0.764

Using the normal distribution table, P = 0.2777

P(X ≤ 400) = 27.77%

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step-by-step explanation: