KZ0.00.10.2The population mean and standard deviation are given below. Find the required probability and determine whether thegiven sample mean would be considered unusual.For a sample of n = 60, find the probability of a sample mean being less than 22.2 if μ = 22 and o=1.3.0.30.4Standard Normal Table (Page 2)0.50.60.70.80.9For a sample of n = 60, the probability of a sample mean being less than 22.2 if µ = 22 and o=1.3 is.(Round to four decimal places as needed.)Would the given sample mean be considered unusual?The sample meanClick the icon to view page 1 of the standard normal table.Click the icon to view page 2 of the standard normal table.Arca0z.05.02 .03 .045080 5120 5160.00 .01.5000 50405398 5438 5478 .5517 5557.5793 583251995596598759105948587162556179 621762936331.6368.6554 6591 6628.6664670067366915 6950 6985.7019.7054.7088 7123.7157 71907224.73577389 .7422 7454754972577580 761176737852.7881 7910 .79397967.7486 .7517.7794 .7823.8078 8106 81338340 8365 838985778212 8238 82648159 81868413 84381.085991.11.21.31.4.9535 .9545.9495 9505 .9515 .9525.9591 .9599 .9608 .9616.9671 9678 .9686.9625 .96339693.9699.9706.9767.7734 77648023 80518289 83158461 8485 8508 8531 855486218643 8665 8686 8708 .8729 .8749 8770.8790 8810 88308849 8869 8888 8907 8925 8944 8962 8980 8997 .90159032 9049 .9066 .9082 .9099 9115 .9131 .9147 9162 .91779192 9207 .9222 .9236 9251 .9265 .9279 .9292 .930693191.5 9332 9345 .9357 .9370 .9382 .9394 .9406 .9418 9429 94411.6 .9452 9463 9474 .94841.7 .9554 9564 9573 .95821.8 .9641 9649 .9656 96641.9 .9713 .9719 9726 .97329738 .9744 .9750 9756.97612.0 .9772 9778 9783 9788 9793 .9798 .9803 .9808 .9812 .98172.1 .9821 .9826 .9830 9834.9838 .9842 .9846 9850 .9854 .98572.2 .9861 9864 .9868 .98719875 .9878 .9881 .9884 .9887 .98902.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 9911 .9913 .99162.4 .9918 9920 .99229925 9927 .9929 9931 .9932 9934 9936.9938 9940 .9941 .9943 9945 .9946 9948 .9949 .9951 9952.9953 .9955 9956.9957 .9959 9960 9961 .99622.7 .9965 .9966 9967.9968.9976 9977 .9977 9978 9979.9982 9983 .9984 .9984 9985.9987 .99879987 9988.9988 .9989 99893.1 99909991.9991 9991 .9992 .9992 9992 .99923.2 9993 .9993 .9994 9994 9994 .9994 .9994 .99953.3 9995 .9995 .9995 9996 .9996 .9996 .9996 .99963.4 99979997 .9997 .9997 .9997 .9997 .9997 .9997.00 .012.52.6.9963 9964.9969 .9970 9971 .9972.997399742.8 .9974 .997599809981.9981 .99822.93.09979.9985.9989Z.02.03.04.05.06.07be considered unusual because it has a probability that is7291 .7324.7642.06.07.085239 5279 5319.5636 .5675 .5714.6026 6064 61036406 6443 64806772 6808 68447704.7995.09.5359.5753614165176879.9986 .9986.9990.99909993.9993.9995.99959996.99979997 .9998.08.09- Xthan 5%. KThe population mean and standard deviation are given below. Find the required probability and determine whether thegiven sample mean would be considered unusual.For a sample of n = 60, find the probability of a sample mean being less than 22.2 if µ = 22 and o=1.3.AreaClick the icon to view page 1 of the standard normal table.Click the icon to view page 2 of the standard normal table.For a sample of n = 60, the probability of a sample mean being less than 22.2 if μ = 22 and o=1.3 is.(Round to four decimal places as needed.)Would the given sample mean be considered unusual?Standard Normal Table (Page 1)02DZ.09.08.07.03.02.01.00.06-3.4 .0002 .0003 0003 .0003.05 .04.0003 0003-3.3 .0003 0004 0004 .0004 0004 0004 0004.0003 0003 .0003 00030005 .0005 0005.0009 .0009 .0009.000700100013.0013 .001300350047-3.2 .0005 .0005 .0005 .0006 0006 0006-3.1 .0007 0007 .0008 .0008 .0008 0008-3.0 .0010 .0010 0011 .0011 .0011 0012 .0012-2.9 .0014 .0014 0015 .0015 .0016 .0016 .0017 .0018 .0018 0019-2.8 .0019 .0020 0021 .0021 .0022 0023 .0023 .0024 .0025 0026-2.7 .0026 .0027 0028 .0029 0030 0031 .0032 .0033 .0034-2.6 .0036 0037 0038 .0039 0040 0041 .0043 0044 .0045-2.5 .0048 0049 0051 .0052 0054 0055 .0057 .0059 .0060 0062-2.4 .0064 .0066 .0068 .0069 .0071 0073 .0075 .0078 .0080 0082-2.3 0084 .0087 0089 .0091 0094 0096 .0099 0102 0104 0107-2.2 01100113 .0116 0119 0122 0125 .0129 0132 0136 0139-2.1 0143 0146 0150 0154 01580162 .0166 0170 .0174 0179-2.0 0183 0188 0192 01970202 0207 0212 0217 0222 0228-1.9 0233 .0239 .02440250 02560262.0268.0274 .0281-1.8 .0294 0301 0307 .0314 0322 0329 .0336 0344 .03510384 0392 0401 0409.0475 0485 0495 05050582 .0594 .0606 06180708 072102870359.04360446-1.7 .0367 .0375-1.6 .0455 .0465-1.5 .0559 .057104270526 .05370548.0643 .0655 .0668.04 18.0516.06.300764 0778 .0793 0808.0968.1093 1112 .1131 11510735 0749-1.4 .0681 0694-1.3-1.208690885.0823 .0838 0853.0985 .1003 .1020 .1038.10561170 .11901210 .1230.1251.10751271.1469 .1492.1711 .1736.1379 .1401 .1423 .1446-1.1.1292 .1314 .1335 1357-1.0.1515 .1539 .1562 1587-0.9 .1611 .1635 .1660 .1685.1762 .1788 .1814 .1841-0.8 .1867 .1894 .1922 .1949 .1977 2005 2033 2061 2090 2119-0.7 2148 21772206 .2236 2266 2296 2327 2358 2389 2420-0.6 2451 2483 2514 2546 2578 2611 .2643 2676 .2709 2743-0.5 2776 2810 2843 .2877 2912 2946 .2981-0.4 3121 .3156 3192 3228 3264 3300 3336-0.3 .3483 35203557.3594 3632 3669.3707-0.2 3859 3897 .3936 .3974 4013 4052 4090-0.14364 4404 4443-0.04761Z.0630853446382142074483 4522 4562 46024920 4960 5000.013015 .3050.3372 .34093745 37834129 41684247 4286 43254641 4681 4721.074801 4840 4880.05 .04.03.09.08.02.00Print.0006 0006 .00070901 .0918 0934 .0951Done- X than 5%.

Answered: Image /qna-images/answer/031b4e79-87a2-445b-a7b0-6bdb66fd3875.jpg

Answered: puinas para The population mean and…

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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Please assist with a,b,c C. Is the p value greater than or lesser than the level of significance? Is there enough evidence to support the claim?
According to one study, brain weights of men are normally distributed with a mean of 1.10 kg and a standard deviation of 0.12 kg. Use the data to answer parts (a) through (e). Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. a. Determine the sampling distribution of the sample mean for samples of size 3. Interpret your answer in terms of the distribution of all possible sample mean brain weights for samples of three men. The mean of the sample mean is µ, kg. (Type an integer or a decimal. Do not round.) The standard deviation of the sample mean is o, = kg. (Round to four decimal places as needed.) Interpret these answers in context. For the brain weights have a normal distribution with a mean of kg and a standard deviation of kg. (Type integers or decimals rounded to four decimal places as needed.) b. Repeat part (a) for samples of size 12. The mean of the sample mean is µ;…
In the 1992 presidential election, Alaska's 40 election districts averaged 1993 votes per district for President Clinton. The standard deviation was 584. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places. a. What is the distribution of X? X ~ N( b. Is 1993 a population mean or a sample mean? Select an answer v C. Find the probability that a randomly selected district had fewer than 2053 votes for President Clinton. d. Find the probability that a randomly selected district had between 2074 and 2208 votes for President Clinton. e. Find the 95th percentile for votes for President Clinton. Round your answer to the nearest whole number.
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The population mean and standard deviation are given below. Find the required probability and determine whether the given sample mean would be considered unusual. For a sample of n = 67, find the probability of a sample mean being less than 22,5 if u= 23 and g = 1.21. E Click the icon to view page 1 of the standard normal table. E Click the icon to view page 2 of the standard normal table. For a sample of n = 67, the probability of a sample mean being less than 22.5 if u = 23 and o =1.21 is (Round to four decimal places as needed.)

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