Question 1203817: Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P31, the 31-percentile. This is the temperature reading separating the bottom 31% from the top 69%.
Answer by ikleyn(52780)   (Show Source):
You can put this solution on YOUR website! .
Assume that the readings at freezing on a batch of thermometers are normally distributed
with a mean of 0°C and a standard deviation of 1.00°C.
A single thermometer is randomly selected and tested. Find P31, the 31-percentile.
This is the temperature reading separating the bottom 31% from the top 69%.
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You are given a normal curve with a mean of 0°C and a standard deviation of 1.00°C.
They want you find a z-score in the number line such that the area on the left
of this z-score under the normal curve is 0.31 of the total area under the normal curve.
For this purpose, you can use standard function invNorm of your regular calculator TI-83/84
area mean SD <<<---=== formatting pattern
P = invNorm(0.31, 0, 1).
You will get the value P = -0.496°C (rouned). ANSWER
Alternatively, you may use free of charge online calculator at this web-site
https://davidmlane.com/hyperstat/z_table.html
It has clearly designed interface, so even a beginner student can learn it and use easily.
Online calculator will give you the same answer.
Solved.
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