August 5, 2017: Outside temperature 71F; Joe’s temperature 62F.
June 2, 2017: Outside temperature 77F; Joe’s temperature 60F.
April 14, 2017: Outside temperature 65F; Joe’s temperature 64F.
February 7, 2017: Outside temperature 40F; Joe’s temperature 61F.
December 11, 2017: Outside temperature 34F; Joe’s temperature 63F.
What is the point? I know you are not referring to me.
Hey Joe. Glad you are here. I have been doing this experiment for some time now. A few years ago, I noticed that Joe’s, the famous coffee shop in New York City is always pleasant – summer or winter. I wanted to see how much variation there is in the internal temperature, so I started measuring it daily. It also meant $5 a day on coffee. 😯
So what did you find at that price?
I discovered that the temperature was discrete and always between 60 and 64. Some of it may be due to the temperature app I use on my phone – it only shows whole numbers; nevertheless, the temperature seems to be pretty regulated. Here is the summary of that data shown in probabilities and a simple probability distribution plot. You must be familiar with this.
Sure, I remember our first conversations about probability. It is the long run relative frequency. So you are finding that the probability that the temperature is 60F on any day is 0.1, the probability that the temperature is 62 on any day is 0.4 and so forth.
Correct. Shall we also compute the summary statistics?
Sure, it will be useful to estimate the expected value (E[X]), variance (V[X]) and the standard deviation of this random variable based on the probability distribution function. Let me do that.
The expected value is
E[X] = 60*0.1 + 61*0.2 + 62*0.4 + 63*0.2 + 64*0.1 = 62
The average temperature is 62F.
Variance can be estimated using the following equation
Applying this to our data: 
So the variance of the temperature is 1.2 deg F*deg F, or the standard deviation is 1.095 deg F.
Fantastic. We can see that the average temperature is 62F and not varying much. That explains why Joe’s is always crowded. Besides having great coffee, they also offer great climate.
I agree. By the way, do you know the geolocation of your readers?
Sure. Here is a map of our readership. We’ve got to get some more countries on board to cover the globe, but this is a great start. Why do you ask?
I ask because outside the USA, people follow the SI unit system and they may want to do these computations in Celsius scale.
Very interesting observation. Do you have a solution for it?
Well, the simplest way is to convert the data first into Celsius scale and re-compute. But that is boring. There’s got to be some elegant approach.
Yes there is. The properties of expected value and the variance operations can help us with this. Going back to your concern, we can convert degree F to degree C using this equation:
In lesson 25, we learned that the expected value operation is linear on functions. For example, 
Using this property, we can compute the expected value of Joe’s temperature in deg C as:
E[C] = 5/9*E[F] - 160/9 = 5/9*62 - 160/9 = 16.66 deg C.
For variance, it is not linear. Let us look at the derivation.
Very clear. Let me apply this to the Celsius equation.
and the standard deviation will be 0.6 deg C.
Excellent. Now that you have your intellectual focus going tell me if the variance of the sum is the sum of the variance for two or more random variables?
Hmm. Going by your previous lessons, your questions always open new concepts. There is something hidden here too that you want me to discover. Can you give me a hint?
Well Joe, Noam Chomsky once said that if you can discover things, you are on your way to being an “independent” thinker.
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