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please help me find average and standard deviation
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- The following molarities were calculated from replicate standardization of a solution: 0.5006, 0.5003, 0.5023, 0.5009, 0.5015, 0.5013, and 0.5022 M. Assuming no determinate errors, what is the upper limit of the 95% confidence interval?What is the relative average deviation in the following percent KHP values in ppt? 46.80% 47.23% 48.03%An analysis of city drinking water for total hardness was done by two students in the laboratory and produced the following results (in ppm CaCO3): Student A: 228.3, 226.4, 226.9, 227.1, and 228.6. Student B: 229.5, 226.1, 230.7, 223.8, and 227.5What is the 95% confidence interval for the mean?
- 7) The homogeneity of the chloride level in a water sample from a lake was tested by an analyst. Below are the concentration in parts per million (ppm) of Cl: 27.32, 27.22, 27.21, 27.24, 27.20 i. Determine the mean and median. ii. Calculate the confidence limit at 90%. iii. Identify a possible outlier and use the Q-test to determine whether it can be retained or rejected at 90% confidence levelDoes the data set of 155.8, 158.4, 194.9, 164.8, 154.5, 184.3, 114.8 and 157.8 ppm contain outlier(s)? (Given that Qcrit.90 % confidence interval (n = 8) D 0.468) %3D %3D O All data should be retained O It is difficult to judge O Both 114.8 and 194.9 ppm are outliers O 114.8 ppm is the outlier O 194.9 ppm is the outlier[References) A method for the determination of the corticosteroid methylprednisolone acetate in solutions obtained from pharmaceutical preparations yielded a mean value of 3.7 mg mL with a standard deviation of 0.3 mg mL. For quality control purposes, the relative uncertainty in the concentration should be no more than 3%. How many samples of each batch should be analyzed to ensure that the relative standard deviation does not exceed 9% at the 95% confidence level? samples
- A group of students titrate a calcium carbonate standard solution repeatedly in five experimental trials. Based on the five trials, the group determines that the average concentration of the standard is 80.5 ppm with a standard deviation of 0.5 ppm. Using the equation for the Student's t confidence interval found in Appendix 1 of your lab manual and the t-value for a 95% confidence interval, calculate the confidence interval for which there is a 95% chance of finding the "true value" of the standard solution's concentration based on the data. Put your answer in the format # lower limit < "true value" < # upper limit. (Appendix 2 has an example that may be helpful.)Propagate the error in the following expression:(0.5847(±0.0002)mole / 250.0(0.1)??) ×(1000 ml / 1 L)Replicate tests showed the following lead content results of a blood sample: 0.752, 0.756, 0.752, 0.751, and 0.760 ppm Pb. Calculate the (a) the mean (b) the standard deviation (c) the variance (d) the relative standard deviation in parts per thousand, (e) the coefficient of variation, and (f) the spread.
- Sample IdentificationCodeConcentration of M%TA=2-log(%T)Q50004.00 x 10-417.90.75Q50013.20 x 10-425.00.6Q50022.40 x 10-435.70.46Q50031.60 x 10-450.20.3Q50048.000 x 10-570.80.15SampleIdentificationCode%TA=2-log(%T)AMQ0210150143.70.359518560.360.000192Q0210150244.10.355561410.360.00018Q0210150343.80.358525890.360.00017Q0210150444.10.355561410.360.00018Q0210150543.80.358525890.360.00017What was their percent error?43%Does Batch 021015 meet legal requirements?No, because it is not between 2.85 * 10(4) and 3.15 * 10(4)Well #DropsBluedye1234567891012345678910Drops 9Distilled water876543210Concne 0.26tration0.52Test Tube #0.781.041.3Solutions3Concentration (M)2.082.32.6Concentration (ppm)1:1 dilution11.82Starting Dilution21.562:1 dilution0A.Zero standard0Was your calibration curve as linear as you expected?B.Did you experience any âdriftâ of the resistance readings?C.What is the equation of your best-fit line?D.What commercial drink did you analyze?E.Assuming…Com X Bb Mas X Mas X Hom X K! Kahx Exan X K! Kah X Mitc X КI Kah X K Kah x K! Kah x Kah x K Kahx C session.masteringchemistry.com/myct/itemView?assignmentProblemID=134214560 Apps New Tab Effect of Salinity on... Sleep Disorders | M... P Philo American Journal ... Zeeshan JBAS Ic Wait for Next Quest... oramge juice Effect ccalculate mean,SD and 95%confidence limit. Use t=4.3.Compare the two means using the student's t test and precision of the two methods using F test at 95% confidence level . For two sets of three replicates use t=2.78 and fcrit=19SEE MORE QUESTIONS