SOLUTION: 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 s
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Question 1207168: 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 the probability of obtaining a reading between -2.572°C and -0.181°C
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the mean is 0 and the standard deviation is 1 degrees
the probabiliy you'll get a reading between -2.572 degrees and -0.181 degrees can be found using the z-score as follows.
z-score formuls is z = (x-m)/s
z is the z-score
x is the critical low score or the critical high score.
m is the mean
s is the standard deviation.
using the low score, the formula becomes z = (-2.572 - 0) / 1 = -2.572.
using the high score, the formula becomes z = -.181 - 0) / 1 = -.181.
you can use a z-score calculator to find that the area to the left of the z-score of -2.572 is equal to .0050556733 and the area to the left of the z-score of -.181 is equal to .428183948.
the area in between is the larger area minus the smaller area equal to .4231281215.
that's the probability that the temperature will be between -2.572 degrees and -.181 degrees centigrade.
some calculators allow you to find it directly, without having to use the z-scorfes.
one such calculator canb e found at https://www.hackmath.net/en/calculator/normal-distribution?mean=3510&sd=25&above=3485&area=below&below=3485&ll=&ul=&outsideLL=&outsideUL=&draw=Calculate
with this calculator, you enter the mean and the standard deviation and the low and high score desired and it tells you the probability.
here are the results.
if you wanted the z-score, rather than the raw score, you would have entered a mean of 0 and a standard deviation of 1.
this problem is unique in that the z-score has a mean of 0 and a standard deviation of 1 and also has a raw score with a mean of 0 and a standard deviation of 1.
that is not usually the case.
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