Supplementary Exercise 4.71 of IPS7e
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A discrete probability distribution for the temperature (X) of the flame in
a glass heating process. As an aside, one might think that a continuous
distribution would be more natural...

The temperature probability distribution:

                   Temperature (degrees F)
               540    545    550     555     560
probability    0.1    0.25   0.3     0.25    0.1 

The probability distribution is valid because the probabilities are >=0 and sum to 1
across the possible outcomes.

(a) 
Mean temperature:
   EX = 540*0.1 + 545*0.25 + 550*0.3 + 555*0.25 + 560*0.1 = 550
The distribution is symmetrical around 550, so we could have guessed
that the mean would equal 550.

We calculate the standard deviation for X in two steps: first the variance, and then
the standard deviation.
   VarX = 0.1*(540-550)^2 + 0.25*(545-550)^2 + 0.3*(550-550)^2 + 0.25*(555-550)^2 + 0.1*(560-550)^2
        = 10+6.25+0+6.25+10 = 32.5
    sdX = sqrt(VarX) = sqrt(32.5) = 5.7

(b)
The new variable of interest is T=X-550 - a translation of the original
variable X by the value (-550). We can either use the rules for a translation, 
or the general rules for linear transformation of a random variable.

    ET = EX-550 = 0 (i.e., we adjust the center by the translation)
   sdT = sdX = 5.7  (i.e., no impact on spread by a translation)

In order to use the formulae from slide 3L-16, you should set b=1 and a=-550.

(c)
Here the variable of interest is Y=1.8*X+32. We should use the mean and standard
deviation rules for linear transformation of random variables (3L-16). 
Note that a=32 and b=1.8.

    EY = 32+1.8*EX = 32+1.8*550 = 1022
   sdY = 1.8*sdX = 1.8*5.7 = 10.26 ~= 10.3.

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Finally, Minitab commands to carry out these calculations after the
distributions has been entered in suitable columns, labeled 'x' and 'p(x)'.

MTB > Name C1 "x"
MTB > Set 'x'
DATA>   1( 540 : 560 / 5 )1
DATA>   End.
MTB > name c2 "p(x)"
MTB > name C3 'meanx'
MTB > Let 'meanx' = 'x'*'p(x)'
MTB > Sum 'meanx'.
Sum of meanx = 550

MTB > name C4 'varx'
MTB > Let 'varx' = ('x'-550)^2*'p(x)'
MTB > Sum 'varx'.
Sum of varx = 32.5

Next a direct calculation of mean and variance for the temperature in
degrees Fahrenheit.

MTB > Name C5 'xF'
MTB > Let 'xF' = 1.8*'x'+32
MTB > Name C6 'meanxF'
MTB > Let 'meanxF' = 'xF'*'p(x)'
MTB > Sum 'meanxF'.
Sum of meanxF = 1022

MTB > Name C7 'varxF'
MTB > Let 'varxF' = ('xF'-1022)^2*'p(x)'
MTB > Sum 'varxF'.
Sum of varxF = 105.3

Note that the Minitab calculator cannot be used to calculate a simple
expression such sqrt(105.3)=10.25 ~=10.3. You would either have to use
the calculator built into Windows or another calculator.