A ginding machine is so set that its production of shafts has an average diameter of 10.10 cms and a standard deviation of 0.20 cms. The product specifications call for shaft diameters between 10.05 cms and 10.20 cms. What proportion of output meets the specifications presuming normal distribution ?
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- A granular soil has an angle of friction with a mean value of 40° and a coefficient of variation of 2.5%. Determine the corresponding mean and standard deviations values for the bearing capacity coefficient, N..A process manufactures ball bearings whose diameters are normally distributed with mean 2.505cm and standard deviation 0.008cm. Specifications call for the diameter to be in the interval 2.5+/- 0.01cm. What proportion of the ball bearings will meet the specification?The existing machine setting of a factory produces bearings with a diameter that is normally distributed with mean and standard deviation equal to 3.0005 cm and 0.0010 cm, respectively. Customer specifications require the bearing diameters to lie in the interval [2.998, 3.002]. Those outside the interval are considered scrap and must be re-machined. With the existing machine selling, what fraction of total production will not be considered scrap?
- The engineer of large manufacturing company takes many measurement of a specific dimension of component from the production line.she finds that distribution of dimension is approximately normal ,with a mean of 3.33cm and a cofficient of variation 3.60percent . 1.what percentage of measurement will be equal to 3.50 cm 2. What value of the dimension will be exceeded by 90 percent of the components ?A product has a nominal length of 11cm. The tolerance associated with this dimension is +0.5cm and -0.5cm. Inspection is carried out on the length and it is found that the mean length is 10.9cm with the standard deviation of 0.47cm. i. Calculate the natural process limits. ii. State the specification limits.iii. Construct process showing all the four limits. In addition, shade the proportion of the product that is outside specification.iv. Calculate both capability indices.The annual rainfall in a certain region is modeled using the normal distribution shown below. The mean of the distribution is 34.1 cm and the standard deviation is 3.4 cm. In the figure, V is a number along the axis and is under the highest part of the curve. And, U and W are numbers along the axis that are each the same distance away from V. Use the empirical rule to choose the best value for the percentage of the area under the curve that is shaded, and find the values of U, V, and W. Percentage of total area shaded: (Choose one) V 25 30 35 40 45 ( cm)
- The annual rainfall in a certain region is modeled using the normal distribution shown below. The mean of the distribution is 36.9 cm and the standard deviation is 4.8 cm. In the figure, V is a number along the axis and is under the highest part of the curve. And, U and W are numbers along the axis that are each the same distance away from V. Use the empirical rule to choose the best value for the percentage of the area under the curve that is shaded, and find the values of U, V, and W. Percentage of total area shaded: (Choos 20 25 30 35 0 40 0 45 50 55 ( cm) Check Save For Later Submit 2021 McGraw Hill LLC. A Rights Reserved. Terms of Use | Privacy Center MacBook 80 F3 88 F4 44 esc F1 F10 FS FB F9 @ 23 $ & 3 5 6 7 8 9 Q E T. Y P A F G H K ock C V alt alt MOSISO command optic ontrol option commandThe annual rainfall in a certain region is modeled using the normal distribution shown below. The mean of the distribution is 41.8 cm and the standard deviation is 5.3 cm. In the figure, V is a number along the axis and is under the highest part of the curve. And, U and W are numbers along the axis that are each the same distance away from V. Use the empirical rule to choose the best value for the percentage of the area under the curve that is shaded, and find the values of U, V, and W. Percentage of total area shaded: (Choose one) 25 30 35 40| 45 50 55 60 ( cm)The annual rainfall in a certain region is modeled using the normal distribution shown below. The mean of the distribution is 36.5 cm and the standard deviation is 5.2 cm. In the figure, V is a number along the axis and is under the highest part of the curve. And, U and W are numbers along the axis that are each the same distance away from V. Use the empirical rule to choose the best value for the percentage of the area under the curve that is shaded, and find the values of U, V, and W. Percentage of total area shaded: (Choose one) ▼ 200 35 | 55 25 30 40 45 50 ( cm) Submit Continue |Privacy Center © 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use DIl S0 FB F7 esc F3
- Find the mean and standard deviation of the thicknessesEstimate the missing data for the station x according to the following information using normal ratio method: Station Normal Annual ppt(cm) A 44.1 B 36.8 C 47.2 X 37.5 O ≈3.847 cm O ≈3.70 cm O ≈3.518 cm ≈3.374 cm ppt(cm) 4.3 3.5 4.8 pxThe standard deviation of measurements of a linear dimension of a mechanical part is 0.14 mm. What sample size is required if the standard error of the mean must be no more than (a) 0.04 mm, (b) 0.02 mm?