EST2703 Fall 1 2018-Dr. Bolton Homework: Week 5 Homework Score: 067 of 1 pt %) 6

Question

EST2703 Fall 1 2018-Dr. Bolton Homework: Week 5 Homework Score: 067 of 1 pt %) 6.2.13-T 4 of 10 (10 complete) > HW Score: 50.33%, 5 03 of 11 Question Help A particular manufacturing design requires a shaft with a diameter between 23 90 mm and 24 012 mm. The manufacturing process yields shafts with diameters normally distributed, with a mean of 24.003 mm and a standard deviation of 0 006 mm. Complete parts (a) through (c) a. For this process what is the proportion of shafts with a diameter botweon 23 90 mm and 24.00 mm? The proportion of shafts with diameter between 23 90 mm and 24.00 mm is 03086 (Round to four decimal places as needed) b. For this process what is the probability that a shaft is acceptable? The probability that a shaft is acceptable is 0.9332 (Round to four decimal places as needed) c. For this process what is the diameter that will be exceeded by only 2 5% of the shafts? | The diameter that will be exceeded by only 2.5% of the shafts is[]mm Round to four decimal places as needed) YS

Explanation / Answer

c) Let the diameter of the shaft that is exceeded by only 2.5% be D

P(X > D) = 0.025

P(X < D) = 1 - 0.025 = 0.975

P(Z < (D - mean)/standard deviation) = 0.975

P(Z < (D - 24.003)/0.006) = 0.975

(D - 24.003)/0.006 = 1.96

(1.96 is the z score corresponding to probability value 0.9750 from standard normal distribution table)

D = 24.0148

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