Part I
- Add your name and date on the above chart using text function and save it as an image on your one-drive. Save this image and submit it with your part II report (5 Marks).
Part II
1.Calculate the probability of observing a cell with 15 pits based on the sample intensity (λ computed from all cells in the image). Write a sentence interpreting what this means. Include commands used to generate the answer. (out of 2)
- 1 mark for the code and correct solution
- 1 mark for the explanation of the method
Expected sentance: with an average of 7.08 event in each unit the probability of seeing exactly 15 event in each unit is equal to 0.003
answer <- c(4,7,9,4,5,0,4,15,9,9,10,3,7,14,16,7,12,7,11,7,15,2,9,3,1,6,14,11,3,1,2,7,8,6,1,6)
dpois(x = 15, lambda = mean(answer))## [1] 0.003637584- What is the probability of observing a cell with between 3 and 5 pits (λ computed from all cells in the image)? Write a sentence interpreting what this means. Include commands used to generate the answer. (out of 3)
- 2 mark for the code and correct solution
- 1 mark for the description
Expected sentance: with an average of 7.08 event in each unit the probability of seeing 3 OR 4 OR 5 events is equal to 0.262
p3 <- dpois(x = 3, lambda = mean(answer))
p4 <- dpois(x = 4, lambda = mean(answer))
p5 <- dpois(x = 5, lambda = mean(answer))
p3+p4+p5## [1] 0.2623633- There are two important assumptions to using the Poisson distribution we should consider.
The probability that an event will occur within a given unit must be the same for all units (i.e. the underlying process governing the phenomenon must be invariant)
The number of events occurring per unit must be independent of the number of events occurring in other units (no interactions/dependencies).
Write short paragraph (200-300 words) explaining why or why not these assumptions are met in this analysis of the weathering pits dataset. (out of 5)
Students should be aware that the effect of location of each cell can play a role in the first assumption. The process of formation of these landscapes are complex and there can be underlying factors there. However since our sample is from one stone some of the processes can be ignored.
The main porpuse of this question is to ask students to think about the as sumptions of probability models before using them
- Explore the assumption of independence by calculating the probability of each observed count and noting where on the image any counts with a probability less than 0.10 occur. Comment on whether these are distributed randomly over the image or clumped in specific parts of the image and what this means for the independence assumption. Include an image showing which cells have an unusual (i.e., p < 0.10) number of pits. (out of 5)
Hint: to answer 4 you can take a screenshot from the lab and use MSPaint (Windows), Photos (Mac) or another graphics program to identify which cells have unusual counts.
1 mark for calculating all the probabilities for all cells 1 mark for making image 3 marks on the discussion on spatial distribution of the pits
prop <- dpois(x = answer, lambda = mean(answer)) # This line is the answer for calculating all the probabilities for all cells
# we can also plor them and show them as a table.
plot(answer, prop, pch=20, cex=2, xlab = "Pit Count", ylab = "Probability")
cbind(prop,answer)## prop answer
## [1,] 0.0880008560 4
## [2,] 0.1489294690 7
## [3,] 0.1037823508 9
## [4,] 0.0880008560 4
## [5,] 0.1246678794 5
## [6,] 0.0008389719 0
## [7,] 0.0880008560 4
## [8,] 0.0036375838 15
## [9,] 0.1037823508 9
## [10,] 0.1037823508 9
## [11,] 0.0735124985 10
## [12,] 0.0496946011 3
## [13,] 0.1489294690 7
## [14,] 0.0077031186 14
## [15,] 0.0016103886 16
## [16,] 0.1489294690 7
## [17,] 0.0279423296 12
## [18,] 0.1489294690 7
## [19,] 0.0473375937 11
## [20,] 0.1489294690 7
## [21,] 0.0036375838 15
## [22,] 0.0210471252 2
## [23,] 0.1037823508 9
## [24,] 0.0496946011 3
## [25,] 0.0059427177 1
## [26,] 0.1471773576 6
## [27,] 0.0077031186 14
## [28,] 0.0473375937 11
## [29,] 0.0496946011 3
## [30,] 0.0059427177 1
## [31,] 0.0210471252 2
## [32,] 0.1489294690 7
## [33,] 0.1318646340 8
## [34,] 0.1471773576 6
## [35,] 0.0059427177 1
## [36,] 0.1471773576 6