Experiment With Voltages of 100 Diodes - Lab Report Example

Assignment 3 - Statistics Task 1 What you measured and how This experiment the measurement of the forward biased voltages of 100 diodes. It was important to note the prevailing temperature it the room at the time of measurement of the voltage of each of the diode. This was important because the diodes are sensitive to temperature. The equipment used to measure the voltages was of high sensitivity with precision of 1mV. Time-series graph Graph showing the voltage of every diode Figure 1 Grouped frequency table Create categories (bins). You should have around 10 bins. Count how many diodes fall into each bin Table 1 Bin Frequency 542 8 546 16 550 24 554 35 558 13 562 1 566 2 570 0 574 0 578 1 More 0 Frequency Histogram – No gaps between the bars Voltage along bottom Frequency up the side Figure 2 Cumulative Frequency Table Add the frequencies together as you go through the groups to find the cumulative frequency. Groups start from zero each time. Table 2 Bin Frequency Cummu f Cumulative % 542 8 8 8.00% 546 16 24 24.00% 550 24 48 48.00% 554 35 83 83.00% 558 13 96 96.00% 562 1 97 97.00% 566 2 99 99.00% 570 0 99 99.00% 574 0 99 99.00% 578 1 100 100.00% More 0 100 100.00% Cumulative Frequency Graph Voltage is plotted on the x-axis, cumulative frequency on the y-axis Points should be joined together with a curve Points should be plotted at the highest point in each group. (Point at 535 for the first group, 540 for the second) Figure 3 Box and Whisker Plot Easiest drawn by hand Only one axis (voltage) Draw a line to show minimum and maximum voltage, median, upper quartile and lower quartile voltages. Figure 4 Task 2 Calculate: Mean (add all voltages, divide by the number of diodes) Median (put voltages in order of size, find the middle value) Both the voltage in the 50 and 51 position are 551 and thus the median is taken as 551 Mode (most common value. It is possible for there to be more than one modal voltage) From the table it can be seen two voltages, 552 and 553 have a frequency of 10 which is the highest frequency. Thus this is the modal value Table 3 Voltage Frequency Percent Valid Percent Cumulative Percent Valid 538.00 1 1.0 1.0 1.0 539.00 1 1.0 1.0 2.0 540.00 1 1.0 1.0 3.0 541.00 3 3.0 3.0 6.0 542.00 2 2.0 2.0 8.0 543.00 1 1.0 1.0 9.0 544.00 4 4.0 4.0 13.0 545.00 5 5.0 5.0 18.0 546.00 6 6.0 6.0 24.0 547.00 5 5.0 5.0 29.0 548.00 2 2.0 2.0 31.0 549.00 8 8.0 8.0 39.0 550.00 9 9.0 9.0 48.0 551.00 9 9.0 9.0 57.0 552.00 10 10.0 10.0 67.0 553.00 10 10.0 10.0 77.0 554.00 6 6.0 6.0 83.0 555.00 9 9.0 9.0 92.0 556.00 3 3.0 3.0 95.0 558.00 1 1.0 1.0 96.0 560.00 1 1.0 1.0 97.0 564.00 2 2.0 2.0 99.0 578.00 1 1.0 1.0 100.0 Total 100 100.0 100.0 Sample standard deviation (NOT population standard deviation). Include the formula and say what each symbol represents. Samples of 10 voltages were taken from the 100 values. To ensure that this was taken randomly the values were picked before any sorting of the data, with the first value to be picked being the one that was number one on the list. The other values were then taken after skipping 9 positions so as to make a sample of 10 values. The values are shown in the column x in the table4 Table 4 x -m (x-)^2 546 -2.8 7.84 555 6.2 38.44 551 2.2 4.84 550 1.2 1.44 550 1.2 1.44 553 4.2 17.64 552 3.2 10.24 546 -2.8 7.84 541 -7.8 60.84 544 -4.8 23.04 5488 173.6 Task 3 – Part 1 - Create a table with your pendulum swing results showing the pendulum length, time for 10 swings, time for 1 swing. Table 5 L=Y TX10 T T2=X 0.3 10.9 1.09 1.1881 -0.8 -3.21183 2.569466 10.31587 0.5 14.2 1.42 2.0164 -0.6 -2.38353 1.43012 5.68123 0.7 16.8 1.68 2.8224 -0.4 -1.57753 0.631013 2.48861 0.9 19.1 1.91 3.6481 -0.2 -0.75183 0.150367 0.565253 1.1 21 2.1 4.41 0 0.010067 0 0.000101 1.3 23.2 2.32 5.3824 0.2 0.982467 0.196493 0.965241 1.5 24 2.4 5.76 0.4 1.360067 0.544027 1.849782 1.7 26.2 2.62 6.8644 0.6 2.464467 1.47868 6.073598 1.9 27.4 2.74 7.5076 0.8 3.107667 2.486134 9.657594 1.1 4.399933 9.4863 37.59728 Calculating gradient m Using the results from the table we have Calculating y-intercept  =1.1-(0.252X4.399933) =-0.0088 Draw a linear graph (straight line graph). Look back through your lesson with Kevin on this. Figure 5 Calculating g=  Calculate the Pearson Product Moment Coefficient to find r. n= number of values (number of pendulum lengths) Y X XY Y2 X2 0.3 0.09 0.027 0.09 0.0081 0.5 0.25 0.125 0.25 0.0625 0.7 0.49 0.343 0.49 0.2401 0.9 0.81 0.729 0.81 0.6561 1.1 1.21 1.331 1.21 1.4641 1.3 1.69 2.197 1.69 2.8561 1.5 2.25 3.375 2.25 5.0625 1.7 2.89 4.913 2.89 8.3521 1.9 3.61 6.859 3.61 13.0321 9.9 13.29 19.899 13.29 31.7337 Task 3 part B Which of the following statements is true for which machine? Statement 1The machine can be expected to mix resin and hardener within the allowed tolerance for all quantities selected. Machine A –The gradient of the line is 1.665 meaning the ratio of resin to hardener is 1 to 1.665 Statement 2The machine mixes hardener and resin but adds a small fixed quantity of additional hardener to all batches. Machine B it has a small Y intercept of 0.063 which is the additional hardener Statement 3The machine mixes hardener and resin but adds a small fixed quantity of additional resin to all batches. Machine A small negative Y intercept which is the additional resin Statement 4This machine adds proportionately too much hardener. Machine B with a gradient of 1.84 it means there the ratio is 1 to 1.84 which is above the required Statement 5This machine adds proportionately too much resin. Machine C the gradient of graph is 1.468 that gives the ratio of 1 to 1.468 which means too much resin Statement 6This machine is the least consistent in measuring chemicals and shows the greatest variation, Machine C Task 4(LO4.4 Use the normal distribution and confidence intervals for estimating reliability and quality of engineering components and systems a) The length of bolts manufactured is believed to follow a normal distribution, given by the equation below. Explain the meaning of the symbols, and plot a graph of probability density given that the mean length of the bolts is 180.6mm with a population standard deviation of 0.35mm. In the equation σ is the standard deviation and  is the mean and x is the actual length of a the bolt Show the ±1σ, 2σ and 3σ points on the graph, and state the confidence intervals normally associated with these lines. Figure 6 b) An engineer is called to service a group of 64 machines. The total working time taken was 48 hours. Determine the 95% confidence interval for the mean service time. Mean time =48/64=0.75 Assuming a standard deviation of 0.1hr then the 95% confidence interval is 95CI = 3σ =0.75+3x0.1=1.05hrs c) A motor has a mean life of 3000 hours, with a standard deviation of 300 hours. Estimate the reliability at the motor at the 2800 hour point. From probability table the probability associated with -0.667=0.2514 This is the probability that the motor will have failed after 2800hrs. Thus the reliability at this point is 1-0.2514=0.75 Read More