can someone please show the steps on how to do these two problems? thank you .

PROBLEM #1

{Exercise 8.51 (Algorithmicn In developing patient appointment schedules, a medical center wants to estimate the mean time that a staff member spends with each patient. How large a sample should be taken if the desired margin of error is two minutes at a 95% level of confidence? How large a sample should be taken for a 99% level of confidence? Use a planning value for the population standard deviation of 7 minutes. 95% Confidence (to the nearest whole number): 99% Confidence (to the nearest whole number): D "I" {I'll-Iv I Ivl' \\vaquI-wl Egg-3:; Emnes' inthelahle givcnemundehc meloielcofthe z value. For example, for z = 1.25, the cumulative probability is .8944. .1!) .01 .02 .03 .04 .05 .06 .07 .08 .09 Z .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 .7580 .761 1 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015 .5032 .9049 .9056 .9082 .9099 .9115 .9131 .9147 .9162 .9177 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441 .9452 .9463 .9474 .9484 .9495 .9505 .95 15 .9525 .9535 .9545 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817 .9821 .9826 .9830 9834 .9838 .9842 .9846 .9850 .9854 .9857 .9861 .9864 .9868 19871 .9875 .9878 .9881 .9884 .9887 .9890 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952 E: P31919198" ."!"'."I" 1'1"??? .... ..... Jinant: eon-Jain lawna coo41ml:- khaki-'6 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 8 8 6 E 8 18 3 E .9989 .9990 .9918} A population has a mean of 400 and a standard deviation of 60. Suppose a sample of size 100 is selected and 53 is used to estimate [1,. Use ztable. a. What is the probability that the sample mean will be within +/- 4 of the population mean (to 4 decimals)? (Round 2 value in intermediate calculations to 2 decimal b. What is the probability that the sample mean will be within +/ 11 of the population mean (to 4 decimals)? (Round 2 value in intermediate calculations to 2 decimal