t study found that 41 percent of shoe soles were contaminated with C. difficile
ID: 3277580 • Letter: T
Question
t study found that 41 percent of shoe soles were contaminated with C. difficile A i As a follow up to this study, researchers sought to further investigate the relation bacteing shoes in the house and household C. diff. levels. The following table gives a two-way classification of 900 randomly selected people on the basis of how much they wear shoes in their home and household C. diff. levels (safe C. diff level vs, unsafe C. diff. level). C. diff. level Shoe wearing in house Safe S Minimal None207 UnsafeV Mt 78 :28 136 260 73.15 Some 134 85 27 A lot If one person is selected at random from this group, find the probability that this person below, provide the appropriate probability rule formula, your work answer with a box around it. Leave your answers in fraction form.) to reflect that formula and your a) (3 points) wears their shoes in their house minimally not tall given that he/she has safe eve so 85 b) (8 points) has safe levels of C. diff. given that he/she wears their shoes in their house minimally not at c) (3 points) has unsafe levels of C. diff. in their house. shoes d) (3 points) wears their shows some or a lot in their house Y5 OD e) (3 points) Is there any indication from the data provided that less wearing of shoes in a house is effective in reducing C. diff. bacteria? Why or why not? Provide a short sentence to justify your answer ys.t ma reduce c,dff acteria,be cause data sliows that s opues frn wearing shoes llo ) (3 points) Are the events "Wears shoes th the house a lot " and "unsafe C. diff. level" independent? Provide a calculation to defend your conclusion. -en Unsate 53+,38-191 tars sho 38Explanation / Answer
below is the table:
a) probability P(A|S) =P(AnS)/P(S) =207/426
b) probability P(S|A) = P(SnA)/P(A)=207/285
c)P(U) =favourable outcome/total outcome =474/900
d) it is ok
e ) it is ok
f) P( wears shoe a lot) =P(C) =85/900 and P( unsafe) =P(U) =474/900
P(C)*P(U) =(85/900)*(474/900) =0.0497
P( wear shoe a lot and unsafe) =P(CnU) =260/900=0.2889
as P(CnU) is not equal to P(C)*P(U) ; thery are not independent
please revert for any clarification required
safe(s) unsafe(U) total minimal/none(A) 207 78 285 Some(B) 134 136 270 A lot(C) 85 260 345 total 426 474 900Related Questions
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