Question

tPacuity 9. (15 pts) What proportion of project completion times fall within two standard deviations of the mean? 10. (15 pts) What proportion of project completion times fall beyond one standard deviation of the mean? 11. (10 pts) 12% of arrivals occur in less than how many days.

Answer 1

X ~ Normal(μ = 35, σ-10.3) Proportion of project completion times fall within two standard deviations of the mean Pụー20 < X < μ + 20) = P(35ー2(10.3) < X < 35 + 2(10.3)) 9. = P(14.4 < X < 556) Let's convert X score into Z score by Z =- 14.4.35<z<ー103 55.6-35 10.3 10.3 = P(-2 < Z < 2) =P(Z < 2) _ P(Z <-2) Since, P(a < Z <b) = P(Z < b)-PCZ <a) =P(Z < 2)-[1-P(Z < 2)] Since, P(Z <-a)-1-P(Z < a) =2*P(Z < 2)-1

TABLE A.2 Cumulative normal distribution (continued) z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 5000 5040 .5080 5120 .5160 5199.5239 5279 5319 5359 0.1 5398 5438 5478 5517 5557 5596 5636 5675 5714 .5753 0.2 5793 5832 5871 5910 .5948 .5987 6026 .6064 .6103 6141 0.3 .61796217 .6255 6293 6331 6368 6406 .6443 6480 6517 0.4 6554 6591 6628 6664 .6700 6736 .6772 6808.6844 6879 0.5 .6915 6950 6985 .7019 .7054 7088 7123 7157 7190 7224 0.6 .7257 729 .7324 .7357 .7389 7422 7454 .7486 7517 7549 0.71.7580|7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 7881 7910 .7939 7967 .7995 8023 805 8078 8106 8133 0.98159 8186 8212 8238 8264 8289 8315 8340 8365 8389 1.0 8413 8438 8461 8485 8508 8531 8554 8577 8599 8621 1.1 8643 8665 8686 8708 8729 8749 8770 8790 8810 8830 1.2 8849 8869 8888 8907 8925 8944 8962.89808997 .9015 1.39032 9049 .9066 9082 .9099 .9115 9131 9147 9162 9177 1.4 9192 207 9222 9236 9251 9265 9279 9292 9306 9319 1.5 9332 9345 9357 9370 9382 9394 9406 9418 9429 9441 1.6 9452 9463 9474 9484 9495 9505 9515 .9525 .9535 9545 1.7 9554 9564 9573 9582 9591 9599 9608 9616 9625 9633 1.8 9641 9649 9656 9664 9671 9678 9686 9693 .9699 9706 1.9 .9713 9719 9726 9732 9738 9744 9750 9756 .9761 9767 2.0 .9772 9778 9783 9788 9793 9798 9803 .9808 .98129817 2.2 9861 9864 9868 9871 9875 9878 9881 9884.9887 .9890 2.3 9893 9896 9898 9901 9904 9906 9909 9911.9913 9916 2.4 9918 9920 9922 9925 9927 9929 9931 9932 9934 9936 From standard normal table, P(Z < 2) 0.9772 Pụー20 < X < μ + 20) = 2(0.9772) _ 1 0.9544 Hence, 95.44% of the project completion times fall within two standard deviations of the mean 10. Proportion of project completion times fall beyond one standard deviation of the mean (i.e below 10 or above 1o) P(X < μ-lo) + P(X > μ + 1σ) = P(X < 35-10.3) + P(X > 35 + 10.3) - P(X< 24.7) + P(X > 45.3)

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P(76.( ))-1-0.12 ))-0.88 using P(X <x) 1P(X<-x) 10.3 a-35 10.3 P(Z <- (2) Now from standard normal table TABLE A.2 Cumulative normal distribution (continued) z 0.00 0.01 0.02 0.030.04 0.05 0.06 0.07 0.08 0.09 0.0 5000 5040 5080 .5120 5160 5199 5239 5279.53195359 0.15398 .5438 5478 5517 5557 5596 5636 .5675 .5714 .5753 0.2 5793 5832 5871 5910 .5948 .5987, 6026 .6064 6103 6141 0.3 6179 6217 6255 6293 6331 6368 6406 6443 6480 6517 0.4 6554 6591 6628 6664 6700 6736 .6772 6808 6844 .6879 0.51.6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 7257 .7291 7324 .7357 7389 7422 7454 .7486 .7517 .7549 0.7 7580 761 7642 .7673 .7704 .7734 7764 7794 7823 .7852 0.8 7881 .7910 .7939 .7967 .7995 8023 8078 8106 8133 0.98159 8186 8212 8238 8264 8289 8315 8340 8365 8389 8621 1.1 8643 8665 8686 8708 8729 874 8770 8790 8810 8830 9015 1.3 9032 9049 9066 9082 9099 9115 9131 9147 9162 9177 1.4 9192 9207 9222 9236 9251 9265 9279 9292 9306 9319 1.0 8413 8438 8461 8485 8508 8531 8554 1.2 8849 8869 8888 8907 8925 8944 8962 For z-1.17, the probability is 0.8790 and for z -1.18, the probability is 0.8810 So we can say for z- P(Z < 1.175) < 0.88 1.171118- 1.175 the probability is 0.88 i.e Comparing (1) and (2) -(10. ) = 1.175 (10.35) - -1175 a 35 10.31.175 a-1.175 10.335 а 22.8975 Hence, 12% of arrivals occurs in less than 22.8975 days