Consider a system with 4 independent stages, each of which must function correctly in succession for the system to operate properly. In other words, stage 1 must function correctly, then stage 2 must function correctly, then stage 3 must function correctly, and then stage 4 must function correctly. If any of the stages fail to function correctly, the system fails. The reliability at initiation (when we attempt to use it) for stage 1 is .9773, stage 2 is .8899, stage 3 is .9913, and stage 4 is .9884. The system reliability at initiation is .8521 . Round your answer to 4 decimal places and do not include a leading zero. You have been given a budget to increase the system reliability as high as possible. You have only two options: 1) you may replace one of the stage's machine with a machine that has a reliability of .9995, or 2) you may add two backup machines to add redundancy to whichever stages you wish one of the backup machines has a reliability of .8500 and the other has a reliability of .8000. If you choose option 1, the maximum system efciency you can achieve is .9570 . Round your answer to 4 decimal places and do not include a leading zero . If you choose option 2, the maximum system eiciency you can achieve is .9954 . Round your answer to 4 decimal places and do not include a leading zero . Your newly optimized system produces an electronic component that fails in an average of 96 months after it is manufactured. The probability that a component will fail in less than 3 years is .9834 . Round your answer to 4 decimal places and do not include a leading zero . You timed every kth dog grooming session over the course of a month, which resulted in 35 timed sessions. There was no pattern regarding the appointment time or day of the week based on the systematic sampling method. You recorded each time in a spreadsheet and calculated the mean to be 39.77 minutes and the standard deviation to be 6.3 minutes. Construct an interval estimate so that you expect the population mean to lie outside of the interval approximately 8% of the time. The standard error is 39.77 , the margin of error is i 1.864 the lower boundary is and the upper boundary is . Round your answers to 4 decimal places. AutoSave OFF Descriptive Statistics Calculator Home Insert Draw Page Layout Formulas Data Review View Tell me Share Comments Times New Roman 11 AA 207 D Wrap Text General Insert v Delete Paste BI UVV MVA Merge & Center v $ ~ % 9 08 Conditional Format Cell Sort & Find & Analyze Sensitivity Formatting as Table Styles Format Q v Filter Select Data x Office Update To keep up-to-date with security updates, fixes, and improvements, choose Check for Updates. Check for Updates 08 Jx B C D E F G H K L M N P Q R S T U V W X Y Z AA AB Data Standardized W N - Sample Population Sample (n) or Population (N) Size Mean Median Mode (Multimode Function) #N/A #N/A #N/A #N/A IN/A #N/A #N/A #N/A #N/A #N/A #N/A #N/A 90 Percent Range for Excel's Sample Skewness IN/A #N/A n Lower 5% Upper 5% n Lower 5% Upper 5% Skewed Left Normal Skewed Right Skewness 0 Trimmed Mean #NUM! #NUM! 20 - 0.84 0.84 90 - 0.41 0.41 Geometric Mean #NUM! #NUM! 30 - 0.69 0.69 100 0.4 0.40 Mid-Range 0.00 0.00 40 - 0.61 0.61 150 0.3 0.33 Mid-Hinge In/a In/a 50 - 0.55 0.55 200 . 0.28 0.28 50 - 0.51 0.51 300 - 0.23 0.23 Standard Deviation #DIV/0! #DIV/O! 70 - 0.47 0.47 100 - 0.20 0.20 Estimated Standard Deviation 0.00 0.00 80 - 0.44 0.44 500 . 0. 18 0.18 Coefficient of Variation (CV) #DIV/0! #DIV/O! Minimum 90 Percent Range for Excel's Sample Kurtosis Coefficient Maximum Lower 5% Upper 5% Lower 5% Upper 5% Platykurtic Mesokurtic Leptokurtic Heavier tallo Normal peak Sharper peak Kurtosis
