As part of an interview for a research position three...

As part of an interview for a research position, three applicants are asked to transfer 150 ?L of distilled water with a P-200 micropipette to a weighing paper and to determine the weight of each drop with an analytical balance. The three measurements made by each of the applicants are listed below. Applicant A: 0.161 g, 0.147 g, 0.142 g Applicant B: 0.158 g, 0.156 g, 0.157 g Applicant C: 0.143 g, 0.153 g, 0.150 g Q1. Calculate and type in the mean and standard deviation of the measurements by each applicant. Q2. Calculate applicants' accuracy of micropipetting, and rank the applicants by accuracy. (A (accuracy) = 100 x Vavg/V0, where A is the accuracy of the pipette, Vavg is the average calculated volume and V0 is the value you set the pipette to dispense. Accuracy should be between 99-101%) Q3. Rank the applicants by their precision of micropipetting. Q4 .Select one of the applicants based on 2. and 3., and explain your selection.

Transcript

00:01 This question is asking us to calculate the mean and standard deviation for each of these students and to relate this to precision and accuracy.

00:12 So our formula for mean is each of our sample values divided by the number of samples we conducted.

00:27 Standard deviation, the formula for this is the square root of the sum of each of our values minus, the mean, that value squared for each of these, and then you divide it by the number of samples minus 1.

00:53 For this question, it's also important to note that the density of water is 1 gram per milliliter.

01:04 So if you draw, say, 150 0 .150 .150, 0 .150 ,050 ,000 mil liters of water, this is equivalent to drawing 0 .150 grams of water.

01:18 So moving on to the calculations for each student, the participant a had values of 0 .161, 0 .147, and 0 .142.

01:36 Participant b had values of 0 .158, 0 .156, and 0 .157.

01:46 Participant c had values of 0 .143, 0 .153, and 0 .153.

01:54 We can add each of these up to get the total and divide it by three because we have three samples for each participant to get our mean value.

02:04 So here we have 0 .161 plus 0 .147 plus 0 .142, which equals 0 .450.

02:16 We can divide 0 .450 by our three samples and we get a mean of 0 .150 grams.

02:27 For participant b, we do the same.

02:30 We add up our values, get our total value, which here equals 0 .471, and 0 .471 divided by our sample amount.

02:48 And we get a mean of 0 .157 grams.

02:58 Here for c, again, same process.

03:07 We get a total of 0 .446 and divided by 3, we get a mean of 0 .149 about.

03:21 This is rounded.

03:22 From there we can calculate our standard deviations.

03:29 So for participant a, we have the square root of our first value, 0 .161, minus the mean of 0 .150, square.

03:47 You add it to our 0 .147 minus 0 .150 squared and 0 .142 minus 0 .15 also squared.

04:04 And we divide this by n minus 1 .1.

04:07 So we divide it by 2.

04:08 When you calculate this out, you get a standard deviation value of 0 .0098 on average.

04:20 We do the same.

04:21 Same for participant b.

04:25 Here we have 0 .158, and i'll just abbreviate this, 0 .156, 0 .157, and we subtract 0 .157, which was the mean, from each of these values, divided by 2, and our standard deviation value for participant b is 0 .001...

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