Question

the shear strength of each of ten test spot welds is determined, yielding the following data (psi): 392 376 401 367 389 362 409 415 358 375 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. 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 m and s? now use the invariance principle.] c. suppose we decide to examine another test spot weld. let x 5 shear strength of the weld. use the given data to obtain the mle of p(x

Answer 1

(a) The sample mean is calculated as 382.3 psi and sample standard deviation is calculated as 19.76 psi.

(b) The 95th percentile is calculated as 414.80 psi that is below 414.80 psi, the 95% of all the welds have their strengths.

(c) There are 81.5% chances that the shear strength of the welds is less than or equal to 400 psi.

What is standard deviation?

A statistic known as the standard deviation, which is calculated as the square root of variance, gauges a dataset's dispersion from its mean. By figuring out how far off from the mean each data point is, the standard deviation can be calculated as the square root of variance.

A higher deviation exists within a data set if the data points are further from the mean; consequently, the higher the standard deviation, the more dispersed the data.

By comparing the annual rate of return of an investment to the standard deviation, a statistical measurement used in finance, one can learn more about the historical volatility of that investment.

Learn more about standard deviation

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