Results Displayed in a Histogram: The histogram has a distinct bell-shaped curve which proves that the weights follow a normal distribution, which now means I have to calculate the mean and the standard deviations of the weights. Process Add up all 80 numbers previously listed above and this leads to the final total weight= 6983 Average the values to find the mean weight. 6983/80 ≈ 87.2875 Next, find the upper quartile(UQ) and lower quartile(LQ) to find the variance LQ= 64 g UQ= 110 g IQR (variance)=46 g The Standard Deviation √46 ≈ 6.78 g In general, there should be about roughly ⅔ (68.3%) of the mass in one standard deviation of the mean, roughly 95.4% between two standard deviations of the mean, and 99.7% between three standard deviations of the mean. However, it is very rare that any value would be more than three standard deviations. …show more content…
51 g and 87.2875 + 6.78= 94.01 g Between 80.51 g and 94.01 g there are exactly 37/80 x 100= 46.25%. This is under value for one standard deviation which leads to speculation about if there was a large harvest during the year creating large apples. 2 standard deviations from the mean is 87.2875 - (2 x 6.78)= 73. 73g and 87.625 + (2x 6.78)= 100.85 g. Between 74.075 g and 101.195 g there are 62 apples which means 62/80 x 100= 77.5%. Again, this value is too low because between two standard deviations there should be 99.7% which leads myself to believe that there wasn’t enough apples collected or a data collection error. 3 standards of deviation from the mean is 87.2875- (3 x 6.78) = 66.95 g and 87.2875 + (3 x 6.78) = 107.63 g Between 66.95 g and 107.63 g there are 77 apples within this range. This means 77/80 x 100= 96.25%. After discovering the percentage was undervalue seeing that at three derivatives it should be at 99.7%, I took the fourth derivative of the apples’
To find the mass percent of acetic acid in vinegar, the molar mass of acetic acid is 60.05 g/mole, and 1.00 g/mol of density, then 0.96 mol×(60.05 g/1 mol) = 57.65 g 57.65 g/1000= 0.0576 ×100 = 5.76% The average mass % of acetic acid in vinegar = 5.53% The average % of acetic acid was 5.53%, which is close to the acidity of the vinegar that was taken in the lab which was 5%, for the different percent we had it could be because of errors in calculations or errors in collecting data. 3.
Calculation: Initial Mass(g)-Final Mass (g)=Change in Mass (g) Trial 1 74.5-62.0=12.5(g) Trial 2 272.7-271.5=1.2(g) Percent Error: 272.7-271.5 x 100 272.7 =0.440% Percent Change: 74.5-62.0 x 100 74.5 (Trial 1) =16.778% 272.7-271.5 x 100 272.7 (Trial 2) =0.440%
gen RHSB=4+(4/(50^.5)) gen out1=cond(meanaRHSA,1,0) gen out2=cond(meanbRHSB,1,0)
To find the density of the three coins, I followed a short and simple procedure. First, I used the triple beam balance to find the mass of each coin to the nearest tenth of a gram. Next, I filled the graduated cylinder with 50 mL of water. Then I put the quarter in the graduated cylinder. In my data table, I recorded the volume of the quarter and the the water in the graduated cylinder.
Application: 1. Find the area under the standard normal curve between z = 0 and z = 1.65. Answer: The value 1.65 may be written as 1.6 to .05, and by locating 1.6 under the column labeled z in the standard normal distribution table (Appendix 2) and then moving to the right of 1.6 until you come under the .05 column, you find the area .450 . This area is expressed as 2.
75.33 grams Weight of the unknown = 0.23 grams Calculation : 75.33-75.10/0.36x100 = 63.8 % recovery Melting point of
3. In this experiment, the percent yield was 90%. This number implies that there was little error in this experiment. However, this result could have been caused by certain external factors.
The percent recovery of the copper was calculated using the equation, percent recovery = (the mass of the copper recovered after all the chemical reactions/the initial mass of the copper) x 100. The amount of copper that was recovered was 0.32 grams and the initial mass of the copper was 0.46 grams. Using the equation, (0.32 grams/0.46 grams) x 100 equaled 69.56%. The amount of copper recovered was slightly over two-thirds of the initial amount.
The average amount of text messages that Betty Smith has used throughout the 12 month period was 1052.833333 messages. 5. What is the average amount of data in MB that Betty used?
My age-adjusted IQ score is 90, which is lower average compared to people my age. The IQ score, 90 is 1 standard deviation away from 100, the average score of all test takers. Standard deviation is a measure of how close or far numbers, in this case scores, are close to the mean. Additionally, it will create a normal bell-curve of the statistics to determine how close the numbers are to the mean, which is 100 for this IQ test.
1 Histogram Tab Per the survey, the data were grouped into 5 different categories (bins), as such the frequency distribution was run for each category and per specific provider for the ease of interpretation of the data presented. The histogram table beside it depicts the graphic distribution of all observations in a quantitative data set to better understand the distribution of the dataset. The height of the column shows the frequency for a specific range of values. AT&T: Results show that categories 1 and 5 are the smallest and that the majority of the data falls into category 3 and 4, which in total represent 73% of the total AT&T users surveyed.
The final product weight for percent yield was only the solid E product, which missed one half of the final product produce. If both products were weight, the percent yield would have been larger that it was. Instead of 22.33%, it could have been 44.66%. To prove that both products were obtained, but only one of the two products was analyze, a TLC plate of the DCM layer, that contains both products, and of the final product, was obtain.
Natural selection isn’t a random process. Natural selection is the major driving force of evolution. By the end of year 3 the number of fruit for the shortest is 59, for the short length it is 41, the long length is 29 and the longest fruit length is 16. The shorter the length the larger amount of fruit is left for that year. This true for all the years, year 0,1,2 and 3 according to the graph.