HW3p1_Statistics2_dusingss

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1120

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Statistics

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Feb 20, 2024

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ENED 1120 – Spring 2024 HW 3.1: Statistics 2 I NDIVIDUAL A SSIGNMENT : See the course syllabus for a definition of what constitutes an individual assignment. Task 1 (of 4) You are asked to determine if it is financially advisable to close a home sales office on Mondays during the month of January. It is assumed that if the sales office is not open on Mondays, customers will not come back and buy a house on another day. You gather data from the past 4 years for January and calculate the following probability of making home sales on a Monday. Sales (n umber of homes) 0 1 2 3 4 5 P(x=S) .80 .10 0.05 0.02 0.02 0.01 a) Determine the following probabilities: Solution Selling any number of houses 20% Selling at least two houses 10% Selling between and including 2 to 4 houses 9% b) What is the expected average number of houses sold on Mondays in January? c) Employees are paid 100% commission which means they have no base salary and only get paid if they sell a house. The commission for each house sold is $10,000. Employees can make $3,000 / day guaranteed if they work at their second job. If the employees are only worried about money, do you predict that employees will want to work in the sales office for the four Mondays in January? Justify your answer? ENED 1120 – HW 3.1 – Spring 2024 Page 1 of 5 .4 houses sold I believe the workers should work on Mondays. Because, if commission is $10,000/house. Let’s just assume this is per employee. Then .4*10,000 = 4000. If they sell that same time all 4 Mondays that is $16,000. That is good amount of money. I think the employees should work those Mondays.
Task 2 (of 4) Complete the diagram and solution parts of the problem presentation mehod for the problems below. Please note in this assignment the axis has been designated for you. In the futrue you might need to designate the axis. Z or x on the axis is very different. Z designates the standard normal curve with mean 0 and standard deviation 1. Therefore 0 is always at the center of the curve and z is the number of standard deviations away from the mean. X is the value of your data. When x is on the axis the mean of x is at the center of the curve. For the first problem (P(z ≥ 2.0)), the figure has been completed to guide you on the expectations for the other problems. Try to make your diagram as accurate as possible. If easier for you, draw the curve on a separate piece of paper with the included area shaded and replace the image in Column 2 with your own image. Find Diagram Solution and how you solved it a) P(z ≥ 2) 2.28%, I looked at the z table and did 1-(z=2) b) P(-0.4 z 0.8) (z=.4) =.66 z=.8 = .79 .79-.66=.13 =13% ENED 1120 – HW 3.1 – Spring 2024 Page 2 of 5 Z Z
c) At what value of z is 0.05 to the right (z=.05) = .52 1-.52 = .48 48% Task 3 (of 4) The amount of soil picked up and moved by each scoop of an excavator is normally distributed, averages 1 cubic yard and has a standard deviation of .07 cubic yards. Find Diagram Solution and how solved a) What is the probability that the excavator picks up between 1.1 and .9 cubic yards the next scoop? Z=(1.1-1)/.07 z=(.9-1)/.07 Z=1.429 = .0763 z=-1.429 = .9236 (.9236-.0763)*100 = 84.73% b) What is the scoop size that is exceeded 95% of the time? (hint: find z value that corresponds to a probability, of > .95 and work backwards) Z = .95 = 1.65 (1+1.65)*.07 = 1.1155 cubic yards ENED 1120 – HW 3.1 – Spring 2024 Page 3 of 5 Z X X
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c) How could you use this information if you were managing a construction site? You can use this information to tell how many scoops it may take to empty a pile. This data could tell you how long it will take to move a pile of dirt for example. Tasks 4 (of 4) File HW3p1Data contains two data sets. The first data set is the length measurements (in cm) of randomly drawn pieces of cut wood from supplier A (lot A). The second data set contains the lengths of cut wood randomly drawn from supplier B (lot B). Determine if the data in each set is close to normally distributed, give reasoning and include any graphs made to support conclusions. Lot A ENED 1120 – HW 3.1 – Spring 2024 Page 4 of 5 For the data lot A using the data statistics we can tell the standard deviation is high at 11.5. This shows that data has a lot of variances. The range of the data is 40 ranging from 80 to 120. The mean sits right in the middle, however. The data for Lot A is below. Lot A Mean 100.0 697 Standard Error 0.810 979 Median 100 Mode 85 Standard Deviation 11.49 761 Sample Variance 132.1 951 Kurtosis - 1.066 58 Skewness 0.017 575 Range 40 Minimum 80 Maximum 120 Sum 20114
Lot B Based on these results do you think the two suppliers have the same type of process? Why? ENED 1120 – HW 3.1 – Spring 2024 Page 5 of 5 For lot be we can see there is similarities to Lot A. The standard deviation is still pretty high at 8.2. The range of this data is 46.88 ranging from 75.5 to 122.5. Since the standard deviation is so high it is not a good data. Lot B Mean 100.2 728 Standard Error 0.580 321 Median 100.6 009 Mode #N/A Standard Deviation 8.227 473 Sample Variance 67.69 132 Kurtosis 0.143 469 Skewness 0.001 922 Range 46.88 854 Minimum 75.46 421 Maximum 122.3 527 No, these two suppliers probably have a different approach. Because the standard deviation of lot B is low than lot A. This means the points are closer to the mean. Lot B’s data is more accurate for the mean. This is a better set of data.