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P(X 10) =0.03+0.08+0.15+0.2+0.19
(b) if 8 and 12 are inclusive
then probability =P(8X12)
(c) probability of no days will be lost =1-P(X 6) =1-(0.03+0.08+0.15+0.2+0.19+0.16+0.1+0.07+0.02)
from above mean =9.79
and std deviation =1.86
here mean represent on average expected number of days construction crew can not work if data is taken over a large number of summer month period.