I didn’t get as far through the unit Random Variables this week as I had hoped I would. However, although I only got through a third of it, I still covered quite a bit of material. I likely spent close to 10 hours working on KA which makes me think that, compared to the first courses I went though, the material I’m going through now are tougher concepts to understand and take longer to wrap my head around. Whereas when I first started KA and everything was more-or-less review, the concepts I’ve been learning about recently seem more abstract and take longer to sink in. All of this is to say the my progress is slower than it was 50 weeks ago which, though it’s a bit disappointing at times, I suppose it makes sense.

The week kicked off by going through exactly what a random variable is, it’s conventional notation, and explaining the difference between a discrete random variable and a continuous random variable:

• Random Variables
  • Maps outcomes to numbers/values (i.e. quantifies outcomes).
  • Typically denoted with a capital letter.
    • Ex. X = {1 if heads, 0 if tails}
    • Y = {0 if heart, 1 if diamond, 2 if spade, 3 if club}
  • In Algebra, variables are often denoted with a lower case letter (ex. x, y, a, b, etc.) which, when solved for, only have one value. A random variable is different in that it may take on many values and if often associated with probability.
  • Discrete VS Continuous
    • Discrete
      • I.e. distinct, separate, countable, finite.
      • A discrete variable will be ‘one-of-x amount’.
      • Ex. year born, heads of tails, number of ants born in the universe tomorrow, etc.
    • Continuous
      • A variable whose measurement is infinite.
      • Ex. EXACT time, or EXACT weight, or number of times the Leafs will lose.
        • Exact time or weight is infinite because, in terms of its precision, the decimal place could technically go on indefinitely whereas the Leafs losing has no limit simply because they suck.

Next I learned about what the expected value of a random variable is and how to calculate it. The expected value of a variable is the sum of each possible outcome multiplied by its probability. It’s worth noting that you can only calculate the expected value of a variable if the variable in question is discrete. Here is an example of how to calculate the expected value:

Number of Workouts Per Week (X)	Probability (P(X))
0	0.1
1	0.15
2	0.4
3	0.25
4	0.1

• Expected value =
  • E(X) = μX = (0 * 0.1) + (1 * 0.15) + (2 * 0.4) + (3 * 0.25) + (4 * 0.1)
    • = 0 + 0.15 + 0.8 + 0.75 + 0.1
    • = 2.1
    • Therefore the expected number workouts done per week equals 2.1.

I then started working through something I still don’t fully understand which was shifting and scaling (i.e. transforming) variables. These questions would often begin by asking me to find the expected value of a variable. Here’s an example question:

• Q. Jon takes two free throws at a carnival game. The following table shows the probability of him making 0, 1, or 2 of the baskets. What’s the expected value of shots he will make the datas’ standard deviation?

Number of shots made (X)	Probability (P(X))
0	0.16
1	0.48
2	0.36

• E(X) = μX = (0 * 0.16) + (1 * 0.48) + (2 * 0.36)
  • = 0 + 0.48 + 0.72
  • = 1.2

(Note: to find the square root I will use “ √ “ which should be viewed as going across the entire equation.)

• S.D.(X) = √ Σ (X_i – μX)^2
  • = √ (0 – 1.2)^2 * 0.16 + (1 – 1.2)^2 * 0.48 + (2 – 1.2)^2 * 0.36
  • = √ -1.2^2 * 0.16 + -0.2^2 * 0.48 + 0.8^2 * 0.36
  • = √ 1.44 * 0.16 + 0.04 * 0.48 + 0.64 * 0.36
  • = √ 0.2304 + 0.0192 + 0.2304
  • = √ 0.48
  • = ~0.69
• Q. If the game costs $15 to play and Jon wins $10 each time he plays, what’s the expected value and standard deviation of the amount of money he could win or lose?

Number of shots made (X)	Probability (P(X))	Money won/lost (Y)
0	0.16	-$15
1	0.48	-$5
2	0.36	+$5

• Once you’ve found E(X) and S.D.(X), you can find E(Y) and S.D.(Y) by using the following equations:
  • E(Y) = μ(Y) = 10 * E(X) – 15
    • =10 * 1.2 – 15
    • = 12 – 15
    • = -3
    • (Note: 10 equals the amount of money he can win per shot and -15 is the amount of money he pays up front. This means that 10 is the scaling factor and -15 is the shifting factor.)
  • S.D.(Y) = 10 * S.D.(X)
    • = 10 * 0.69
    • = 6.9

I still don’t fully understand how these equations work. I think they’re the same as the equations I would use when I would transform shapes and functions so I don’t think it will take too long before I figure it out. I think a big part of why I don’t completely get it has a lot to do simply with my lack of understanding about expected values and standard deviations.

The last thing I learned this week was about combining random variables. This is done by taking two independent random variables (often called distributions which I think of as any type of probability graph such as the Normal Distribution) and either adding or subtracting their expected values and/or variances. The equations for all four types of combinations are:

	Mean (μ)	Variance (S.D.^2)
Addition	μZ = μX + μY	S.D.^2_(Z) = S.D.^2_(X) + S.D.^2_(Y)
Subtraction	μZ = μX – μY	S.D.^2_(Z) = S.D.^2_(X) + S.D.^2_(Y)

When adding or subtracting the means of X and Y you simply add or subtract them together as you would with any number to get the mean of the new variable, Z. When adding the variances of X and Y you again simply add them together to get the variance of Z, however for some reason when subtracting the variance of Y from the variance of X you still add them together. I’m not exactly sure why this is the case. As a side note, I’ve realized that I don’t have a great understanding of either standard deviation or variance. I know their formulas but I don’t understand why their formulas are the way they are or how someone came up with them.

It’s hard to believe I’m going into Week 52. As I mentioned last week, I’m pretty stoked that I’ve kept at this for so long. This coming week I’d like to get through the majority of the unit Random Variables (720/2100 M.P.) and maybe even finish it off. After this unit, there are only 7 units left in the course; 3 of them are fairly small at 700, 800, and 700 M.P. apiece and 3 others don’t have any Mastery Points available at all (which is the first time that’s ever happened). I’m hoping this means I’ll be able to get through the remainder of the course relatively quickly, at least by the end of November.

As a final note, I found out this week that I may be able to tutor at the tennis club I work at this year! It would be a great way for me to make extra money while also being a way for me to revise what I’ve been working on. I also feel like it would be a strong and logical step towards using math in some type of career. I’m going to speak with one of the managers about it tomorrow so hopefully I’ll have a positive update for the blog next week!