A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 18 subjects had a mean wake time of 101.0 min. After treatment, the 18 subjects had a mean wake time of 80.4 min and a standard deviation of 21.8 min. Assume that the 18 sample values appear to be from a normally distributed population and construct a 95% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 101.0 min before the treatment? Does the drug appear to be effective? Construct the 95% confidence interval estimate of the mean wake time for a population with the treatment. 69.2 min < < 91.5 min (Round to one decimal place as needed.) What does the result suggest about the mean wake time of 101.0 min before the treatment? Does the drug appear to be effective? The confidence interval includes drug treatment does not have an effect. the mean wake time of 101.0 min before the treatment, so the means before and after the treatment could be the same. This result suggests that the 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 .8980 .8997. .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 1.5 .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 .9812 .9817 2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857 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 2.5 9938 .9940 9941 9943 .9945 .9946 .9948 .9949 .9951 9952 2.6 9953 .9955 .9956 .9957 .9959 .9960 .9961 9962 .9963 9964 2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 9973 .9974 2.8 9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981 2.9 9981 .9982 .9982 .9983 .9984 .9984 .9985 9985 9986 .9986 3.0 .9987 9987 .9987 9988 .9988 .9989 .9989 .9989 9990 .9990 3.1 9990 .9991 .9991 .9991 .9992 .9992 .9992 9992 .9993 .9993 3.2 .9993 .9993 .9994 9994 .9994 .9994 .9994 .9995 .9995 .9995 3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 9996 .9997 3.4 9997 .9997 9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998 3.50 9999 and up NOTE: For values of z above 3.49, use 0.9999 for the area. "Use these common values that result from interpolation: z score Area 1.645 0.9500 2.575 0.9950 Common Critical Value Confidence Critical Level Value 0.90 1.645 0.95 1.96 0.99 2.575 Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The aircraft can carry 40 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6,440 lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is 6,440 lb greater than 40 =161 lb. What is the probability that the aircraft is overloaded? Should the pilot take any action to correct for an overloaded aircraft? Assume that weights of men are normally distributed with a mean of 174.4 lb and a standard deviation of 39.6. The probability is approximately . (Round to four decimal places as needed.) Should the pilot take any action to correct for an overloaded aircraft? OA. Yes. Because the probability is high, the pilot should take action by somehow reducing the weight of the aircraft. OB. No. Because the probability is high, the aircraft is safe to fly with its current load. Find the indicated critical value. 20.04 Click to view page 1 of the table. Click to view page 2 of the table 20.04= (Round to two decimal places as needed.) Standard Normal Table (Page 2) Standard Normal (z) Distribution: Cumulative Area from the LEFT N .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 5000 .5040 .5080 .5120 .5160 5199 .5239 .5279 .5319 0.1 .5398 .5359 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 0.2 .5753 .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 0.5 .6879 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 3.7389 .7422 .7454 .7486 .7517 .7549 0,7 .7580 .7611 .7642 .7673 .7704 7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 8078 .8106 .8133 0.9 .8159 .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 .8980 .8997 .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 1.5 .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 .9812 ,9817 2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857 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 2.5 9938 .9940 .9941 .9943 .9945 9946 19948 .9949 .9951 .9952 2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964 2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 9973 .9974 2.8 9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981 Print Done Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. The area of the shaded region is (Round to four decimal places as needed.) 98 Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is between -2.19 and 3.72 and draw a sketch of the region. Sketch the region. Choose the correct graph below. OA. 2.19 3.72 The probability is (Round to four decimal places as needed.) B. -2.19 3.72 C. -2.19 3.72 G O D. a -2.19 3.72 Q G