Biomedical Instrumentation Systems
1st Edition
ISBN: 9781133478294
Author: Chatterjee
Publisher: Cengage
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Categorize each as a determinate or an indeterminate error and further classify determinate errors as instrumental, method or personal.d) The tip of the pipet used in the analysis is broken. e) In measuring the same peak heights of a chromatogram, two technician each report different heights.
Provide an appropriate response.
12) The IQ scores of adults are normally distributed with a mean of 100 and a standard
deviation of 15. Find the probability that a randomly selected adult has an IQ score
between 85 and 125.
The following are the results for the % Cu of ore sample:
24.52%, 24.75%, 24.82%, 24.69%, 24.78%
Calculate the mean.
(A) 24.71%
B) 0.12%
24.75%
0.47%
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- Suppose a normally distributed set of data with 8900 observations has a mean of 123 and a standard deviation of 16. Use the 68-95-99.7 Rule to determine the number of observations in the data set expected to be above a value of 107. Round your answer to the nearest whole value. Hint: This problem is asking for how many observations not the percent. Answer Tip: Don't round any probabilities or percentages in your calculations. Keep all decimal places and round at the END of the problem.arrow_forward:0.29 µg Cu/mL (s → ơ). The analysis of an oil from a reciprocating aircraft An atomic absorption method for determination of copper in fuel samples yielded a pooled standard deviation of spooled engine showed a copper content of 8.03 ug Cu/mL. Calculate the 95 and 99% confidence intervals for the result if it was based on (This problem requires values in your textbook's specific appendices, which you can access through the OWLV2 MindTap Reader. You should not use the OWLV2 References' Tables to answer this question as the values will not match.) a a single analysis 95% confidence interval = 8.03 + µg Cu/mL 99% confidence interval = 8.03 ± ug Cu/mL b the mean of 5 analyses 95% confidence interval = 8.03 + | ug Cu/mL 99% confidence interval = 8.03 ± µg Cu/mL c the mean of 15 analyses 95% confidence interval = 8.03 + ug Cu/mL 99% confidence interval = 8.03 ± ug Cu/mLarrow_forwardPlease answer D : data attached if neededarrow_forward
- A series of standards were analyzed which gave the following results SOLUTION % Transmittance 0 mL of 0.100 mg/mL Hg + 2 mL tinchloride in a total volume of 75 mL 100 1.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 81.3 2.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 63.9 3.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 55 5.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 35.1 How would you make a Beer-Lambert Plot with this information?arrow_forwardStaple your graph to the data sheet. Complete the following table: T-1(K-1) In(VP) 353 -0.00297 348 - o 06191 343 =0.00296 338 333=0.0300 323-0.0030u る= =0.00309 323 273=0.006 0.00366arrow_forwardCategorize each as a determinate or an indeterminate error and further classify determinate errors as instrumental, method or personal. a) An unknown sample being weighed is hygroscopic. b) One component of a mixture being analyzed quantitatively by GC reacts with the column packing. c) A radioactive sample being counted repeatedly without any change in condition yields a slightly different count at each trial. d) The tip of the pipet used in the analysis is broken. e) In measuring the same peak heights of a chromatogram, two technician each report different heights.arrow_forward
- A peak elutes from an HPLC column 13.8 cm in length in 14.8 min. What would be the width at half-height of the peak (in seconds) if the plate height were 7.75 µm? W 1/2 4.1 W 1/2 Incorrect What would be the width at half-height of the peak (in seconds) if the plate height were 46.0 µm? 24.3 Incorrect S Sarrow_forwardMy System Suitability Tests can contribute to PQ True Falsearrow_forwardThe detection limit is the lowest measurable concentration True Falsearrow_forward
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