Assume that the readings on the thermometers are normally distributed with a mea

Question

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 P97, the 97th percentile. This is the temperature reading separating the bottom 97% from the top 3% Click to view page1 of the table. Click to view page 2 of the table Which graph represents Pg7? Choose the correct graph below The temperature for Poz is approximately°. (Round to two decimal places as needed.) Standard Normal Table (Page 1) NEGATIVE z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT 04 06 -3.50 -3.4 and oo01 00 0oo5 006 o08 oo10015 0021 0027 0036 lower 003 oo05 006 o009 012 16 0021 28 007 048 24 0062 0045 00330023 006 0on 0oog 00os 0004 00o2 -24/007 0136 0170 0212 0262 0322 oss2 0475/05n ·0681 00030003 33 07 o09 oo13 001 023 0030 0039 05 66 0084 0003 0002 0005 0005 004 0004 0004 0004 0004 0004 0003 0006 0006 0009 0009 0009 0008 0008 0008 0008 007 007 20 0179 0136 0102 075 oss 0040 0029 0021 001400o 190228 0174 0132 0099 073 0os 003g 0028 0020 0014 -18/0287 0222 0170 0129 .0096 .00 0052 .0038 .0027 -O079 oom 0023 23 0022 0021 0 0039 29 0013 0oog 0006 00040003 -2.9 0016 0016 0025 0024 00450044 0080 0078 200260018 0013 000g oos 000403 0047 0043 0041 2.5 0062 0060 0071 0069 ,0066 0064 0087 0084 0n3 -2.2 -2.310139 OTA 027 0268 0329 ,001 O485 0582 9A 0104 0102 0099 0096 0094 0091 0119 .0793 .0643 0576 .0409 ,0322 2s0192 / .046 .070 0764 06495 0392 03070239 0183 0116 0150 0162 .0158 0207 0202 0274 0268 0262 .0256 0250 .0244 0166 0197 -19 0359 A60526 0630 0749 0294 03920384 03750367 0465 0455 0559 0301 Click to 0446 035) 027212 01620122 0091 0o68 0049 0o36 0436 0427 0418 0516 0409 0401 0505 0495 0548 1.5 0606 0594 0582 0571 .0 0808 0708 0694

Explanation / Answer

From the given z-table, z-value representing the area of 0.03 in the left tail of the graph is -1.88

Here we want to know 97th percentile which means this value would be in the right of the curve. Hence z-value = 1.88

Using central limit theorem,
xbar = mu + z*sigma
xbar = 0 + 1.88*1
xbar = 1.88

Graph : Option B

P(97) = 1.88

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