Q 5.The thickness (in millimeters) of the coating applied to disk drives is one characteristic thatdetermines the usefulness of the product. When no unusual circumstances are present, the thickness (z) hasa normal distribution with a mean of 5mm and a standard deviation of 0.1mm. Suppose that the processwill be monitored by selecting a random sample of 30 drives from each shift's production and determining 2,the mean coating thickness for the sample. (Round your answer to 4 decimal places if any.)(b) When no unusual circumstances are present, we expect 2 to be within 30, of 5mm, the desired value.An z value farther from 5 than 30, is interpreted as an indication of a problem that needs attention.Compute 5+ 30,(c) Referring to Part (b), what is the probability that a sample mean will be outside 5+30, just by chance(that is, when there are no unusual circumstances)?(d) Suppose that a machine used to apply the coating is out of adjustment, resulting in a mean coatingthickness of 4.95mm. What is the probability that a problem will be detected when the next sample istaken?Q(b) is not needed and it is given as a reference jusst solve for Qc) and Qd).iwill be very appreciate!!-0.4-0.3AppendixSTANDARD NORMAL DISTRIBUTION: Table Values Represent AREA to the LEFT of the Z. score.J02Z.00.01.05-3.900005 00005 00004 0000400004.06 007.0900004 00004 00003 .00003.00005.00008-3.800007 00007 000070000600006 .00005 00005000090000800008 0000800011 00010 00010 0001000016 00015 00015 0001400023 00022 00022 0002100034 00032 00031 0003000048 00047 00045 0004300069 00066 00064 0006200058 00056 00054 00052-3.1 00097 00094 00090 .0008700082 00079 00076 00074-3.0 00135 00131 00126 0012200118 00114 00111 00107 0010400187 00181 00175 .00169 00164 00159 00154 .00149 0014400256 00248 00240 .00233 .00226 00219 00212 00205 .0019900347 00336 00326 00317 00307 00298 00289 00280 00272-2.6 00466 00453 00440 00427 00415 00402 00391 00379 00368-2.5 00621 00604 00587 00570 00554 00539 00:23 00508 0049400820 00798 00776 0075500695 00676 0065701072 01044 01017 00990 0096401390 01355 01321012550178601618-2.0 02275 02222 02169 02118 02068-1.9 02872 02807 02743 02680 02619-1.8 03593 03515 03438 03362 03288-1.7 04457 04363 04272 04182 040930128701743 01700 0165900734 0071400939 00914 00889 0086601222 01191 01160 0113001578 01539 01500 0146302018 01970 01923 018760238503074 0300503836 0375401831023300293804746 0464805821 0570502559 0250003216 0314404006 0392003673-1.6 05480 05370 05262 05155 05050 04947 0484604551-1.5 06681 06552 06426 06301 06178 06057 0591805592-1.4 08076 07927 07780 07636 07493 07353 07215 07078 06944 0681109680 09510 09342 09176 09012 08851 08691 08534 08379 0822611507 11314 11123 10935 10749 -10565 .10383 .10204 10027 0985313567 13350 13136 12924 12714 12507 12302 12100119001170215866 15625 .15386 15151 -14917 -14686 14457 14231 14007 13786-0.2-0.1-0.0-2.3-0.9 .18406 .18141-0.8 21186 20897 20611-0.7 .2419623885-0.6 2742527093-0.5 30854 30503 30153-1.2-1.1.17879 .1761920327235762676334458 .34090 3372438209 37828 3744842074 41683 4129446017 45620 4522450000 49601 49202.04000040000600009000140001300020 00019.00029 .0002800042000600008417361200450001300012 0001200019 00018 0001700027 00026 0002500040 00039 00038 0003602442.0001100017.00024.00035.00050.0007100100.00139.0019300264.00357.00480.006390084217106 .16853.16109.19766 .19489186732327022965.22663 223632147626435261092451029806 29460277603120725785 2546329116.32636 32276 31918.3631740129.4403848803 48405 48006 4760835942 .355693519733360 3299737070 3669340905 4051744828 444333482739743 .39358 389743859143644 43251 42858 4246547210 46812 464140110101426.16602 .1635419215 .1894322065 21770248252514328774 28434 2809631561 STANDARD NORMAL DISTRIBUTION: Table Values Represent AREA to the LEFT of the Z score.Z.00.01.02.090.0 500000.1 539830.2 579260.30.450399 5079854380 5477658317 58706.61791 6217265542 .65910662760.5 .69146 .69497 69847.73237.764240.67257575804.72907.761150.70.878814 .79103 793890.981594 .81859838911.19544996080 9616496327.96712.96856 96926.9699597062.96784974419738197500 97558 9761597670979329798298030 98077 9812498169.03.04.05.06.07.0851197 51595 $1994 52392 52790 53188 5358655172 55567 55962 56356 56749 57142 5753559095 59483 59871 .60257 .60642 .61026 .6140962552 .62930 .63307 .63683 .64058 .64431 .64803 .65173.66640 .67003 .67364 .67724 68082 68439 .68793.70194 .70540 .70884 .71226 .71566 .71904 .72240.73565 .73891 74215 .74537 .74857 75175 .75490.76730 .77035 77337 .77637 77935 .78230 .78524.79673 .79955 80234 .805118078581057 81327.82121 .82381 .82639 .82894 .83147 .83398 .836461.0 84134 .84375 .84614 .84849 85083 .85314 .85543 .85769 85993 .86214.86433 .86650 86864 .87076 .87286 .87493 .87698 .87900 .88100 882981.2 .88493 88686 88877 .89065 .89251 .89435 .89617 .89796 89973 .901471.390320 90490 90658 .90824 .90988 91149 .91309 91466 91621 917741.4 91924 92073 92220 92364 92507 92647 92785 92922 93056 931891.5 93319 93448 .93574 .93699 93822 .93943 94062 94179 94295 944081.6 94520 94630 94738 94845 94950 95053 95154 95254 953521.7 95543 95637 95728 95818 .95907 95994.962461.8 96407 96485 96562 .966381.9 97128 97193 97257 973202.0 97725 97778 97831 9788298257 98300 .983419838298645 98679 .98713 .9874598956 98983 .99010 .99036 .99061 .99086 99111 .9913499202 99224 99245.99266 99286 99305 99324 9934399413 .9943099446 99461 99477 9949299560 99573 99585 .99598 99609 9962199674 .99683 9969399702 99711 9972099760 99767 99774 99781 99788 997959981999825 .99831 99836 99841 99846 99851 998563.0 .99865 99869 99874 .99878 99882 .99886 99889 998933.1 99903 .99906 99910 .99913 .99916 .99918 .99921 999243.2 9993199934 99936 99938 9994099942 .99944 999463.2 .99931 .99934 .99936 .99938 .99940 .99942 .999443.3 99952 99953 .99955 99957 99958 .99960 999613.4 .99966 99968 99969 .99970 .99971 .9997299978 .99979 99980.99985 .99986 .99986.99990 .99990 9999199993 .99994 .9999499996 99996 99996985742.1 982142.2 9861098422 .98461 98500.9853798778 98809 98840 98870988992.3 98928991582.4 .99180993612.5 9937999506.995202.6 .995349963299643.99396.9954799664997522.7 99653.997282.8 99744998012.9 9981399896.999469996299973 .9997499981 .99981 .99982.99937 .99988 .99988.99987.9999199992 .99992 99992.99994 99994 .99995 99995.99996 99996 99996 999973349253.5 .99977 .99978.99985.999903.6 .999843.7 .999893.8 99993999933.9 .99995 999959992699948997369980799861.999009992999950.99948.99950.9996499965.999759997699983 9998399989999929999599997

Given Information: Sample size n=30 Mean μ=5 Standard deviation σ=0.1

Answered: (e) Referring to Part (b), what is the…

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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23. Telephone calls from an office are monitored and found to have a mean duration of 452 s and a standard deviation of 123 s. Determine (a) (b) the probability of the length of a call being between 300 s and 450 s. the proportion of calls likely to last for more than 700 s.
The amount of paint required to paint a surface with and area of 50m2 is normally distributed with mean 6L and standard deviation 0.3L. (a). if 6.2L of paint is available what is the probability that the entire surface can be painted? (b). how much paint is needed so that the probability is 0.9 that the entire surface can be painted? (c). what must the standard deviation be so that the probability is 0.9 that 6.2L of paint will be sufficient to paint the entire surface?
"Given the information provided, what is the probability of obtaining a sample mean within +/- 2 of the population mean? Population information: mean = 375, and standard deviation = 15. Sample size is 300."
5. The inside diameter of a randomly selected piston ring is a random variable with mean value of 15cm and standard deviation 0.05cm.
11. The amount of time that a drive-through bank teller spends on a customer is a random variable with a mean μ = 6.4 minutes and a standard deviation o=3.5 minutes. If a random sample of 49 customers is observed, find the probability that their mean time at the teller's window is (a) at most 5.6 minutes; (b) more than 7.4 minutes; (c) at least 6.4 minutes but less than 7.0 minutes. Click here to view page 1 of the standard normal distribution table.¹ Click here to view page 2 of the standard normal distribution table.² (a) The probability that the mean time is at most 5.6 minutes is (Round to four decimal places as needed.) (b) The probability that the mean time is more than 7.4 minutes is (Round to four decimal places as needed.) (c) The probability that the mean time is between 6.4 minutes and 7.0 minutes is (Round to four decimal places as needed.)
3. The mass of coffee in a randomly chosen jar sold by a certain company may be taken to have a normal distribution with means 203 g and standard deviation 2.5g. (i) Find the probability that a randomly selected jar will contain at least 200g of coffee. Obtain the probability that two randomly selected jars will together contain between 400g and 405g of coffee. (iii) The random variable ī denotes the mean mass (in grams) of coffee per jar in a random sample of 20 jars. Identify the value of a such that P(T – 203| < a) = 0.95. (ii)
5. The sediment density (g/cm³) from water in the Red River flowing through Winnipeg, Manitoba has a distribution with mean value 2.65 and standard deviation 0.85. If 40 random samples of water are collected, what is the approximate probability that the average sample sediment density is between 2.65 and 3.00? (Retain two decimal places in all calculations.) Mean value :2.45 40 samples. P (z.u5sx<3.w)?
The melting point of a binder used in manufacturing a rocket propellant is approximately normally distributed, the mean is thought to be 100, and the standard deviation is 4. you wish to test HO: u=100 versus HI: uz100 with a sample of n= 16 specimens. suppose that the true mean melting point is 102. What is the probability of type ll error (B) for the two tail.test with a=0.05? Oa. 0,121999 b.0.4879 30.2439 Od.0.0615 Clear my.choICO
The time taken by employees at Grace Floral shop to put a bouquet together has a normal distribution with mean 26.4 minutes and standard deviation 8 minutes. (i) Find the probability that an employee chosen at random takes between 24.6 and 27.8 minutes to put a bouquet together. (ii) 12% of employees take more than t minutes to put a bouquet together.Find the value of t.
17. A 20,000-piece manufacturing run of 2200 resistors has a mean value of 2200 (naturally!) and a standard deviation of 2.50. They are to be sold as a 2% tolerance product, which means that the values must be between 0.98 x220 N = 215.60 and 1.02 x 220 0 = 224.40.
17) Suppose we want to test H0 : μ ≥ 30 versus H1 : μ < 30. If we do not know the populationstandard deviation, but have a sample mean of 27 and sample standard deviation of 4, for asignificance level of 0.05, what is the value of the test statistic for a sample size of 100?
4. A sample of 100 fluorescent light tubes from the Short Life Tube Company has a mean life of 20.5 hours and a standard deviation of 1.6 hours. Test: (a) At the 1% level whether the sample comes from a population with mean 23.2 hours. (b) At the 5% level whether it comes from a population with mean 20.8 hours. (c) At the 5% level whether it comes from a population with mean less than 20.8 hours

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