Question

4. The number X of days in the summer months that a construction crew cannot work because of the weather has the probability distribution 6 0.03 7 0.08 8 0.15 9 0.20 10 0.19 11 0.16 12 010 13 0.07 14 002 Find the probability that no more than ten days will be lost next summer. Find the probability that from 8 to 12 days will be lost next summer Find the probability that no days at all will be lost next summer Compute the mean and standard deviation of Interpret the mean in the context of the problem Seven thousand lottery tickets are sold for $5 each One ticket will win $2,000, two tickets will win $750 each, and five tickets will win $100 each Let X denote the net gain from the purchase of a randomly selected ticket. a. Construct the probability distribution of X b. Compute the expected value E(X) of x. Interpret it's meaning. C. Compute the standard deviation o of X

Answer 1

4) Answer: 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 (8=X<=12) -P(X-8)+PCX-9)+P(X=10)+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) =0.0

d) ( X-μ)P (x) p(x) x2P(x) (x-u)2 ΧΡ(x) X 0.431 6 0.030 0.180 1.080 14.364 7 0.080 0.560 3.920 7.784 0.623 9.600 0.150 1.200 3.204 0.481 0.200 1.800 16.200 0.624 0.125 1.900 19.000 10 0.190 0.044 0.008 11 0.160 1.760 19.360 1.464 0.234 1.200 12 0.100 14.400 4.884 0.488 0.910 13 0.070 11.830 10.304 0.721 14 0.020 0.280 3.920 17.724 0.354 02 total 99.310 1 9.79 60.397 3.4659 std deviation= Vo2= 1.8617

From the above table mean =9.79
And standard deviation =1.86

• Mean represent on average expected number of days construction crew cannot work if data is taken over a large number of summer month period.