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Dima020 [189]
3 years ago
14

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.)
Mathematics
1 answer:
dybincka [34]3 years ago
8 0

Answer:

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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Answer:

P(X>400)=P(\frac{X-\mu}{\sigma}>\frac{400-\mu}{\sigma})=P(Z>\frac{400-440}{20})=P(z>-2)

And we can find this probability using the complement rule:

P(z>-2)=1-P(z

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(z>-2)=1-P(z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

X \sim N(440,20)  

Where \mu=440 and \sigma=20

We are interested on this probability

P(X>440)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>400)=P(\frac{X-\mu}{\sigma}>\frac{400-\mu}{\sigma})=P(Z>\frac{400-440}{20})=P(z>-2)

And we can find this probability using the complement rule:

P(z>-2)=1-P(z

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(z>-2)=1-P(z

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