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

Question

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. Let  represent the reading of this thermometer at freezing. What reading separates the highest 11.07% from the rest? That is, if ,

find c.   °C

Answer 1

So let's digest some of the information that we have been given.

First, Let's identify the key words for this problem. We have Normally Distributed, Mean, and Standard Deviation.

Second, Let's identify what these words mean and how they are related to one another.

-Normally Distributed. This means that whenever we take a measurement the Probability that we will get a number different from the mean drops off as a gaussian distribution, In math this means we are pulling from the Probability Density Function (pdf): f(x) = (1/(σ√2*π))*exp(—(x-µ)²/(2σ²))

where exp represents the exponential and π = 3.14159...

-Mean. This is just that µ symbol in the pdf and represents the most likely value. (while this is unimportant here it may come in handy later)

-Standard Deviation. This is the σ value in the pdf and is a measure of how spread out our function is.

Whew! That is a lot of information, now we need to figure out how to use it! In order to get the probability from the pdf we must take the integral of our pdf. The probability between any two values in the pdf is given by:

P(a < x < b) = ∫_a^b pdf dx

From the problem statement we have x = z in this case, a = c, and b = ∞ (since we want everything higher than c, if this is not the case hopefully it is clear how to adjust). Therefore, we want:

P(c < z < ∞) = ∫_c^∞ f(z) dz

where f(z) is the normal pdf above and µ = 0, σ = 1.

Finally we know what the probability should be! So we plug in that value and solve for c

0.1107 = ∫_c^∞1/(√2π)*exp(—z²/2) dz = ∫₀^∞~ dz - ∫₀^c~dz = 1/2 * (1-erf(c)) the ~ means to repeat what was written before, namely the pdf

and viola! We have done it!

Now you might ask what next? We have to start guessing values for c and numerically evaluating using what is called the Error function erf(x). You can make educated guesses though to help reduce the time necessary to solve this problem such as it must be greater than 1σ away (because this only captures 68% of the probability). You can also plot the error function for this integral (using wolfram alpha or a similar tool) and find where it is equal to 0.1107.

Hope that this has helped. Good luck! and remember, never stop learning