Anticipated consumer demand in a restaurant for free range steaks next month can be modeled by a normal random variable with mean 1,300 pounds and standard deviation 90 pounds. a. What is the probability that demand will exceed 1,100 pounds? b. What is the probability that demand will be between 1,200 and 1,400 pounds? c. The probability is 0.10 that demand will be more than how many pounds? E Click the icon to view the standard normal table of the cumulative distribution function. O Standard Normal Distribution Table a. The probability that demand will exceed 1,100 pounds is 0.01 0.02 0.03 0.04 0.05 0.06 0.00 0.08 0.09 0.5040 0.5438 (Round to four decimal places as needed.) 00 0.5000 0.5080 05120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5478 0.5517 0.5557 0.5596 05636 0.5675 0.5714 0.5753 0.2 05793 0.5832 0.5871 0.9910 0.9948 0.5987 0.6026 06064 0.6100 0.6141 03 06179 06217 0.6255 0.6293 0.6331 0.6368 0.6406 06443 0.6480 0.6517 04 0.6554 06991 0.6628 0.6664 0.6700 0.6736 0.6772 06808 0.6844 0.6879 0.5 06915 06950 0.6985 0.7019 0.7054 0.7088 0.7123 07157 07190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7580 07881 07 0.7611 0.7642 0.7673 0.7704 07734 0.7764 0.7794 0.7823 0.7852 0.8 07910 07939 0.7967 0.7995 0.8023 O8051 08106 08133 09 08159 O8186 08212 0.8238 0.8264 08289 0.8315 08340 08365 0.8389 1.0 08413 08438 0.8461 0.8485 0.8508 08531 0.8554 0.8577 08599 08621 1.1 08643 O665 08686 0.8729 08749 0.8770 08790 08810 0.8830 1.2 08849 O8888 0.8907 0.8925 0.8944 0.8962 0.80 08997 0.9015 1.3 0.9032 09049 0.9066 0.9082 0.9099 09115 0.9131 09147 09162 0.9177 14 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 15 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 09429 0.9441 16 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0964 0.9573 0.9582 0.9591 09599 0.9608 0.9616 0.9625 0.9633 18 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 09690 0.9706 19 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 0.9854 0.9887 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9857 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9890 23 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 24 0.9918 0.9920 0.9922 0.9925 0.997 0.9929 0.9931 0.9932 0.9934 0.9936 25 0.998 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 26 0.9953 09956 0.9957 0.9999 0.990 0.9961 0.9963 0.9964
Anticipated consumer demand in a restaurant for free range steaks next month can be modeled by a normal random variable with mean 1,300 pounds and standard deviation 90 pounds. a. What is the probability that demand will exceed 1,100 pounds? b. What is the probability that demand will be between 1,200 and 1,400 pounds? c. The probability is 0.10 that demand will be more than how many pounds? E Click the icon to view the standard normal table of the cumulative distribution function. O Standard Normal Distribution Table a. The probability that demand will exceed 1,100 pounds is 0.01 0.02 0.03 0.04 0.05 0.06 0.00 0.08 0.09 0.5040 0.5438 (Round to four decimal places as needed.) 00 0.5000 0.5080 05120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5478 0.5517 0.5557 0.5596 05636 0.5675 0.5714 0.5753 0.2 05793 0.5832 0.5871 0.9910 0.9948 0.5987 0.6026 06064 0.6100 0.6141 03 06179 06217 0.6255 0.6293 0.6331 0.6368 0.6406 06443 0.6480 0.6517 04 0.6554 06991 0.6628 0.6664 0.6700 0.6736 0.6772 06808 0.6844 0.6879 0.5 06915 06950 0.6985 0.7019 0.7054 0.7088 0.7123 07157 07190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7580 07881 07 0.7611 0.7642 0.7673 0.7704 07734 0.7764 0.7794 0.7823 0.7852 0.8 07910 07939 0.7967 0.7995 0.8023 O8051 08106 08133 09 08159 O8186 08212 0.8238 0.8264 08289 0.8315 08340 08365 0.8389 1.0 08413 08438 0.8461 0.8485 0.8508 08531 0.8554 0.8577 08599 08621 1.1 08643 O665 08686 0.8729 08749 0.8770 08790 08810 0.8830 1.2 08849 O8888 0.8907 0.8925 0.8944 0.8962 0.80 08997 0.9015 1.3 0.9032 09049 0.9066 0.9082 0.9099 09115 0.9131 09147 09162 0.9177 14 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 15 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 09429 0.9441 16 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0964 0.9573 0.9582 0.9591 09599 0.9608 0.9616 0.9625 0.9633 18 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 09690 0.9706 19 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 0.9854 0.9887 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9857 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9890 23 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 24 0.9918 0.9920 0.9922 0.9925 0.997 0.9929 0.9931 0.9932 0.9934 0.9936 25 0.998 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 26 0.9953 09956 0.9957 0.9999 0.990 0.9961 0.9963 0.9964
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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