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A particular manufacturing design requires a shaft with a diameter of 21.000 mm, but shafts with diameters between 20.988 mm and 21.012 mm are acceptable. The manufacturing process yields shafts with diameters normally distributed, with a mean of 21.004 mm and a standard deviation of 0.004 mm. Complete parts (a) through (d) below. a. For this process, what is the proportion of shafts with a diameter between 20.988 mm and 21.000 mm? The proportion of shafts with diameter between 20.988 mm and 21.000 mm is 0.1586. (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 0.9772. (Round to four decimal places as needed.) c. For this process, what is the diameter that will be exceeded by only 5% of the shafts? The diameter that will be exceeded by only 5% of the shafts is 21.0106 mm. (Round to four decimal places as needed.) d. What would be your answers to parts (a) through (c) if the standard deviation of the shaft diameters were 0.003 mm? If the standard deviation is 0.003 mm, the proportion of shafts with diameter between 20.988 mm and 21.000 mm is 0.0917 (Round to four decimal places as needed.) If the standard deviation is 0.003 mm, the probability that a shaft is acceptable is 0.9962 (Round to four decimal places as needed.) If the standard deviation is 0.003 mm, the diameter that will be exceeded by only 5% of the shafts is mm. (Round to four decimal places as needed.) - a Screen Shot 2022-07-05 at 6.55.08 PM Given a normal distribution with = 100 and 10, complete parts (a) through (d). Click here to view page 1 of the cumulative standardized normal distribution table. Click here to view page 2 of the cumulative standardized normal distribution table. a. What is the probability that X> 80? The probability that X>80 is 0.9772 (Round to four decimal places as needed.) b. What is the probability that X<75? The probability that X<75 is 0.0062 (Round to four decimal places as needed.) c. What is the probability that X<90 or X> 125? The probability that X<90 or X> 125 is 0.1649. (Round to four decimal places as needed.) d. 80% of the values are between what two X-values (symmetrically distributed around the mean)? 80% of the values are greater than and less than (Round to two decimal places as needed.)