Question

Problem 6: A universal tensile testing machine was used to determine the ultimate tensile strength (k psi) of an unknown alloy steel. The 10 samples had the following ultimate tensile strength in ksi of: 248, 255, 258, 253, 253, 252, 250, 252, 255, 256. Determine the mean, variance, and 99.9% confidence interval assuming normal distribution.

Answer 1

Given 10 samples are as follows: 248, 255, 258, 253,253, 252, 250, 252, 255,256 Mean (1) = sum of samples Total number of samples l - (248 +255+258+253+253 +252 +250 +2 52 - +255+256 lo Til = 253-2 Variance (o?) = (Xi – u)? _ iai Here, N-10 -=[248 – 253-2) +(855–253.257 ...... + (256-253-2)] a or=[45-2) + (1-8)= (4.80+ €0.5%+ 60.27" +62)+(-3-2)7 41.7+(1.87'+6-8)]

Standard deviation (0) = 17-76 = 2-78 (-=2.78 From normal distribution table, our required probability 0.999 is achieved when -3.99 <Zc3-99 > -3.99 < XM < 3.99 > -3.990 < x-l c 3.990 7 63.990+ll) < x < (3.990+4) = (-3-99(2-78) +283-2) < x < (3-992:78) +253-2) → 242-12 < X <264.29