The number X of days in the summer months that a construction crew are cannot work because of the weather has the probability distribution X P(x) 6 0.03 7 0.08 8 0.15 9 0.20 10 0.19 11 0.16 12. 0.10 13 0.07 14 0.02 A. Find the probability that no more than ten days will be lost next summer. B. Find the probability that from 8 to 12 will lost next summer. C. Find the probability that no days at all will be lost next summer. D. Compute the mean and standard deviation of X. Interpret the mean in the context of the problem.
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Solution:
(a)
P(X 10) =0.03+0.08+0.15+0.2+0.19
=0.65
(b) if 8 and 12 are inclusive
then probability =P(8X12)
=P(X=8)+P(X=9)+P(X=10)+P(X=11)+P(X=12)
=0.15+0.2+0.19+0.16+0.10
=0.80
(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)
=1-1
=0
(d)
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.