TABLE 1 TABLE 2 Negative z Scores Standard Normal (z) Distribution: Cumulative A
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TABLE 1
TABLE 2
Negative z Scores Standard Normal (z) Distribution: Cumulative Area from the Left z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 negative 3.50 and lower 0.0001 -3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 -3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003 -3.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005 -3.1 0.001 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007 -3 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.001 0.001 -2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 -2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.002 0.0019 -2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.003 0.0029 0.0028 0.0027 0.0026 -2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.004 0.0039 0.0038 0.0037 0.0036 -2.5 0.0062 0.006 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048 -2.4 0.0082 0.008 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 -2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084 -2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.011 -2.1 0.0179 0.0174 0.017 0.0166 0.0162 0.0158 0.0154 0.015 0.0146 0.0143 -2 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 -1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.025 0.0244 0.0239 0.0233 -1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 -1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 -1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 -1.5 0.0668 0.0655 0.0643 0.063 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 -1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681 -1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 -1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.102 0.1003 0.0985 -1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.123 0.121 0.119 0.117 -1 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379 -0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.166 0.1635 0.1611 -0.8 0.2119 0.209 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 -0.7 0.242 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148 -0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 -0.5 0.3085 0.305 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.281 0.2776 -0.4 0.3446 0.3409 0.3372 0.3336 0.33 0.3264 0.3228 0.3192 0.3156 0.3121 -0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.352 0.3483 -0.2 0.4207 0.4168 0.4129 0.409 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 -0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 0 0.5 0.496 0.492 0.488 0.484 0.4801 0.4761 0.4721 0.4681 0.4641 Note: For values of z below -3.49, use 0.0001 for the area. *Use these common values that result from interpolation: z score Area -1.645 0.05 -2.575 0.005 Assume that the readings on the thermometers are normally distributed with a mean of 0 and standard deviation of 1.00 C. A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to P the 97th percentile. This is the temperature reading separating the bottom 97% from the 97 top 3%. Click to view age 1 of the table. Click to view page 2 of the table Which graph represents P Choose the correct graph below. O D. O A. O B. O C. The temperature for P is approximately 97 (Round to two decimal places as needed.)Explanation / Answer
Here we have to find P97 or 97 percentile.Whenever we go further in percentile it is always from left side so we can reject (A) and (D) .
similarly, it is we are calculating percentile so we can't use Graph (b) as it is telling just a value above 97% percentile.
so option C is correct, which tell about all the 97 % values in the given graph.
Here P97 = 0.97 probabiltiy of occuring an equal or less that the certain temeprature.
Here mea temperature = 0o C andStandard devaition = 1o C
so look into the table for p - value 0.9700 and we found it is in between Z = + 1.88 and Z = + 1.89
and more closer to 1.88 so we will take Z = 1.88
so P97 = + Z * = 0 + 1.88 * 1 = 1.88o C
Here we get 97% percentile value and its graph.
Thank YOu!!!
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