Question

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

389

405

409

367

358

415

376

375

367

362

(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		382.3  psi
standard deviation		20.8 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.)

(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).]

 

 

Answer 1

Concepts and reason

The concepts of maximum likelihood estimator for normal distribution is used in this problem.

Maximum likelihood estimator is the method that gives the values of the one or more parameters for the provided statistics that maximizes the likelihood of the given observations.

Fundamentals

For normal distribution the maximum likelihood of the population mean is the sample mean and of the population standard deviation is the sample standard deviation which is the unbiased estimator of it.

The sample mean is defined as,

The sum of square is calculated as,

Expectation of the square of the deviation of the random variable X from its mean is called variance. It can be calculated using the following formula,

Percentile is defined as a number where a certain percentages of scores fall below that percentile.

95^th percentile is the value below which 95 % of the data lies.

Normal distribution is continuous probability distribution with parameters and.

The continuous random variable X follows Normal distribution,

Mean is and variance is.The continuous random variable X should be converted to standard normal variate

Z is standard normal variate with. The probability of Z is calculated by using the excel function.

(a)

The sample mean is calculated as,

The Sum of Square deviation (SS) is calculated as,

The variance is calculated as,

The standard deviation is calculated as,

(b)

The strength value below which 95% of all the welds will have their strength is calculated as,

(c)

Let X be the random variable that denotes the shear strength of the weld. The probability that shear strength of the weld is less than or equal to 400 is calculated as,

The probability is calculated in excel and is shown below:

Ans: Part a

The true average shear strength is 382.3 psi and standard deviation of the shear strength is 19.76 psi.

Part b

The strength value below which 95% of all the welds have their strengths is 414.80 psi.

Part c

The probability that shear strength of the welds is at most 400 is 0.815.