AU23 Project 2

Project 2 Lillian Delgado For all problems, show your calculations or your work in Excel and explain your thinking. There is a video included in this week’s learning which can be used to help with your calculations. 1. A college professor distributed a bag of M&Ms to each student in their class. Your bag is shown below. a) Calculate the percentage of each color and show your calculations in the table. Write your answer as a decimal and round to 3 decimal places. (Example: 7/33 = 0.212 is how that should look). To calculate the percentage, I used the following formula: - (frequency of given color/Total frequency) × 100 5+6+7+8+10+18 = 54 Color Percentage Brown (5/54) x 100= 9.259% Red (6/54) x 100= 11.111% Blue (7/54) x 100= 12.983% Green (8/54) x 100 = 14.815% Yellow: (10/54) x 100 = 18.519% Orange (18/54) x 100 = 33.333% The professor asked students to submit their M&M totals in a class poll. The data is below. Note: 6 students who were in class did not submit their totals in time to be counted (maybe they were busy eating the M&Ms). b) Calculate the mean and median for the total M&Ms using the 20 numbers above. Show your calculations or add a screenshot of your work in Excel. 50+55+53+55+51.5+49+51+51+51+52+54+47.5+50+53= 723 Mean 724 14 Mean equals = 51.64 The mean number of M&Ms would be the sum of all the M&Ms divided by the total number of students: =723 = (6x50) = 1,023 Therefore, the mean number of M&Ms would increase from 51.64 to 51.15 Color Percentage Brown 9.259% Red 11.111% Blue 12.983% Green 14.815% Yellow 18.519% Orange 33.333% 50 55 53 55 51.5 49 51 51 51 52 54 47.5 50 53 0 0 0 0 0 0

Project 2 c) First, we need to calculate the total number of M&Ms for all the students, excluding those who didn't submit their total: Please see question number one. d) Which measure for the average should the professor use to describe the average number of M&Ms? Why? the median only considers the middle value(s) of a dataset and is not affected by the values of the other data points. 51.55 e) Suppose the 6 students who didn’t submit their total in class all had 50 M&Ms in their bag. How would that change the mean number of M&Ms? How would that change the median number of M&Ms? Write a sentence to explain your thinking. If the 6 students who didn't submit their total in class had 50 M&Ms in their bag, the mean number of M&Ms would change as follows: 50+55+53+55+51.5+49+51+51+51+52+54+47.5+50+53= 723 Mean 724 14 Mean equals = 51.64 The mean number of M&Ms would be the sum of all the M&Ms divided by the total number of students: 723 = (6x50) = 1,023 Therefore, the mean number of M&Ms would increase from 51.64 to 51.15 The median number of M&Ms, however, would remain the same because the middle value is still 51.5, even if we assume that the missing students all had 50 M&Ms. This is because the median only considers the middle value(s) of a dataset and is not affected by the values of the other data points. In addition to collecting total numbers, the professor collected counts for each color. The totals are below. Show your calculations for each color in the table, rounded to 3 decimal places. Qty 86 Brown Qty 88 Red Qty 286 Blue Qty 137 Green Qty 106 Yellow Qty 320 Orange (86/54) x 100 = 159.259% f) Compare the percentages from your bag to the percentages for the entire class. What changes do you notice? Which colors stayed the most consistent? Brown: Class 9.259% vs. Bag 159.259% Color Percentage Brown 159.259% Red 162.963% Blue 529.630% Green 253.704% Yellow 196.296% Orange 592.593%

Project 2 If you counted 32 orange M&Ms in a sample from the bag, you could estimate the total number of M&Ms in the entire bag based on the proportion of orange M&Ms in your sample. h) Suppose a bag of M&Ms is dumped on the table and you count 32 orange pieces. Approximately how many M&Ms do you expect to be in the entire bag based on the percentages of the entire class? Write a sentence to explain your thinking and your calculations. Calculation: 0.011852 (proportion) * 2700 M&Ms (the total number of M&Ms in the bag), which gives you an estimated total of approximately 31.989 M&Ms. Proportion of orange M&Ms in your sample: (32 orange M&Ms / 2700 M&Ms) ≈ 0.011852. This proportion represents the fraction of the bag that your sample represents. The percentages of M&M colors from the class and the fact that you found 32 orange M&Ms in your sample, you can estimate that there are approximately 31.989 M&Ms in the entire 5-pound bag. This estimation is based on the proportion of orange M&Ms in your sample and the class percentages for the various M&M colors. Based on the class percentages and counting 32 orange M&Ms in a sample, we would expect approximately 31.989 M&Ms in the entire 5-pound bag. Estimated total number of M&Ms = 0.011852 x 2,700 M&Ms = 31.989 M&Ms. So, based on the percentages of the entire class and the 32 orange M&Ms in your sample, you would expect approximately 31.989 M&Ms in the entire 5-pound bag