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E Homework: 12.5 Homework Question A normal distribution has mean 60.7 and standard deviation 2.25. Find the data value corresponding to z = - 3.14,

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E Homework: 12.5 Homework Question A normal distribution has mean 60.7 and standard deviation 2.25. Find the data value corresponding to z = - 3.14, The data value is (Round to the nearest tenth as needed.) Help me solve this View an example Get more help .Homework: 12.5 Homework Question 21, 12.5.6 A normal distribution has mean 66.3 and standard deviation 4.52. Find the data value corresponding to the following value of z. Z= 1.23 The data value corresponding to z = 1.23 is (Round to the nearest tenth as needed.) Help me solve this View an example Get more help -Homework: 12.5 Homework * Question 19, 12.5.49 HW Score: 59.09%, 13 of 22 points Q Points: 0 of 1 the moan dotting time of blood in 7 45 seconds with a standard deviation of 30 seconds, What is the probability that an individual's clotting time wil be keys than 2 seconds or greater than 8 seconds? Assume a moul disinbuton The probability is Help me solve this Whew an example Gel more help . FinalchePeter Davis teaches a course in marketing. He uses the Grade Score in Class system shown to the right for assigning grades to his A Greater than x + 1.3s students. From the information in the table, what B x +0.4s to x + 1.3s percent of the students receive the grade B? C x - 0.4s to x + 0.4s Click here to see page 1 of the table of areas. D x - 1.3s to x - 0.4s Click here to see page 2 of the table of areas. F Below x - 1.3s From the information in the table. % percent of the students will receive the grade B. (Round to the nearest tenth as needed.) Help me solve this View an example Get more help * 64"F Light rain\f0.2 oz., the probability of a box being undomusicbil h as needed } X Standard Normal Curve Areas (page 1) The column under A gives ine proportion of the area under the entire curve that is and a negative value of I can be botwoon 2= 0 and a positive value of z. found by using the conosponding poodu vallup of 2. A A A A A 0.00 10.000 0,30 0.118 0.60 0.226 0.90 0.316 120 0 385 1.50 0430 0.01 0/004 031 0.122 061 0.220 091 0319 121 1.51 0 02 0.37 0.126 0.62 0232 0 321 122 030 1.52 0.436 0.03 0.012 0 33 0 129 0.63 0 236 0.324 1.23 0 301 153 0 437 0.010 0.34 0.133 0 64 0.239 124 151 0.05 0.35 0.137 0 65 0 242 0.329 1.55 0.00 10.024 10.141 0 65 10 245 0.90 0 331 1.26 1 56 OOT 0.028 0 37 0.144 137 157 0 148 0 252 0 335 0 400 1.58 0 032 1 38 01030 0 30 0.255 0 330 1 20 0401 150 0-45 0 010 0 70 0 341 1.30 0.159 131 1.61 0.17 0. 163 0 72 0340 1 37 0 160 1 03 1 33 013 8410 0.14 0351 Clear al 04ST we solve this View an example 0.13 0-45 0 355 0413 0 47 in an Print Donoorth as needed. ) Standard Normal Curve Areas (page 1) 0.07 0.028 0.37 0.144 0.67 0 249 10 97 0.334 127 0.348 0 442 0.08 0.032 0 38 0.148 0.68 0.252 1 98 0 336 1 28 0.09 0.036 0.39 0.152 0.69 10.255 0 339 1.29 0401 1.59 OM4 0.10 0 040 0.40 0.155 0.70 0 258 100 0.341 1.30 1.60 0 445 0.11 0.044 0.41 0.159 0.71 0 261 1.01 0 344 1.31 0405 1.61 0446 0.12 0.048 0.42 0. 163 0.72 0.264 1 02 0.346 1 32 0.407 1.62 0 447 0.13 0 052 0.43 0 168 0.73 0.267 1.03 0 348 1.33 0.408 1 63 0.14 0 056 04 0.170 074 0 270 1.04 0351 134 0 410 1.64 0449 0.15 101080 0.45 0174 0 75 0 273 1.05 0.353 1.35 0411 1.65 10 451 0.16 01064 0 46 0.1TT 0.76 0276 1.06 0.355 1.36 0 413 1.60 0.452 DIT 0 181 O.TT 0 279 1.07 137 0415 0 453 0071 0 184 0 78 0.282 0:380 0 416 0454 0 10 DOT5 0.108 0.79 1.09 0 362 1 30 0454 10.20 0.079 150 0 191 0 80 5 788 1.10 0 418 1.70 0155 021 0 083 0.190 0 81 0.291 141 0 421 0 52 1 12 142 0422 1.72 0 22 0. 1 012:94 0 091 1-13 0 371 1 43 10424 0.23 6205 10 300 01373 10 438 924 0 209 0 85 1 15 0 103 0 212 0 3FF sip me solve this View an example 0210 0 28 10.110 0 58 0210 OUT 1 18 0:341 1. PA 6232 0313 Print Donai5 02 oz, the probability of a box being un dance writh as needed } X Standard Normal Curve Areas (page 2) A I A A A A A 1 80 0.484 2.10 0.482 240 0.482 2.70 0.497 3.00 3.30 0 500 1.81 0.485 211 0.483 241 0 492 271 0.497 3.01 10429 3.31 0.500 1.82 2.12 0.483 2.42 0.492 272 0.497 3102 3.12 0.500 1 83 0 468 2.13 0 483 2.43 0 402 273 0 497 3.31 0:500 0 500 184 0.467 214 0484 2.44 274 0.497 3104 0498 3.34 185 0.408 2.15 0.484 2.45 0493 275 305 3 35 0500 0 500 1 BB 0 469 2 18 0.485 248 0.403 2 76 10.497 306 0 459 3 30 217 0.485 247 277 0.497 307 0459 0500 0.409 1 88 0470 2 18 0485 248 0493 2 78 309 0499 3 38 0500 2.10 249 0404 2.70 300 0 499 1.30 1.89 0.471 1:500 1.90 0471 230 0.480 250 2 80 3. 10 0 499 151 0 480 251 10:434 0 498 341 3 42 1 02 0473 0 487 252 3 12 1.03 2 21 253 7 83 0 408 1 04 254 3.14 344 0 474 1 05 235 2:35 285 10 405 10 498 3.10 208 10 501 Clear add D ATT 2:50 0:300 me solve this View an example 3.20 350 321 351 201 231 10 400 201 in and Print Donaw do your x E Result List: al x ProQuest tbo x MyLab Math / X * Homework Hi X P Do Homework X M Inbox (540) - X earson.com/Student/PlayerHomework.aspxThomeworkld=618931259Biquestionld=48flushed=falseadd=6849827&back=http:/mylab.pearson.com/Student/DoAssignments.as. Maps * How can I cancel m... \\\\ Paraphrasing Tool | E Alzheimer's disease. o Abheimer's diete o Probability and Statistics 1773 62 Kimber Carmen Carmen Martinez 0.2 oz , the probability of a box being andnow In as needed. X Standard Normal Curve Areas (page 2) 1.87 0 469 2.17 0 485 2 47 01 493 2.77 307 0 409 0 500 0.470 2 18 0:485 2 48 0.403 278 01500 0 471 2.19 0.404 279 3 39 0471 2 20 2.50 0404 2 80 3 10 0-453 3:40 0 412 221 2151 0 404 3 41 341 01500 1 02 0 473 2 22 0 404 3.12 050 D.413 2 23 2 53 0404 283 313 343 0509 104 0 474 234 0 4AT 254 0 404 314 2 25 10.488 2155 0495 205 3.16 345 1.05 0 475 236 0 483 250 10495 3.16 01500 DATO 327 257 10 495 J IT 3 AT 1.08 0 470 0 489 250 10405 318 0500 DATT 250 3.10 2 0 D ATT 2:30 351 0 500 2.01 DATS 231 261 291 321 202 0 478 2:32 303 0 400 203 264 134 204 205 3.25 3105 Clear all 2 08 10 480 View an example G 0401 2 AT 327 solve this 358 10500 0 481 2 38 0491 3.20 750 0500 2 09 Print Done

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