I’m once again disappointed with how this past week went. I failed the Random Variables unit test 3 times this week (a.k.a. didn’t manage to get 100%). My scores got progressively better after each attempt which was somewhat redeeming I suppose. Because there are 21 questions on this test, there’s a lot of pressure which makes me second guess my answers. If I get a question wrong I have to redo its corresponding exercises and what makes it worse is that I have to get all the remaining questions correct otherwise I have to do their exercises, as well. What’s most frustrating is that often the few questions I get wrong is because of some silly miscomprehension about the question, not that I’m unable to do the math. The cherry on top of all of this is that my allergies are still bad which has only added to the frustration I’ve been feeling this week. AAAAAGH!!!!!!!!!!

… Ok. Moving on…

Considering I only did the unit test, I don’t actually have any new material to talk about here. Even though I’ve already wrote about everything I’ve learned in this unit in previous blogs, I figure I’ll list the types of questions that come up on this test and explain how each type of question typically works. Hopefully this will help me to better understand each of their concepts.

1) Bernoulli Distributions

• Is used when looking at the probability of only 2 possible outcomes.
• Each outcome is labelled as either ‘success’ or ‘failure’.
• The ‘successful’ outcome is assigned a value of 1 and the ‘failure’ outcome is assigned a value 0.
• For example, if the ‘successful’ outcome has a probability of 0.25 and the ‘failure’ outcome has a probability of 0.75, the way to calculate the expected value would be:
  • E(V) = μX = (0 * 0.75) + (1* 0.25)
    • = 0.25
• In that example, if there are 8 trials, the formula would be:
  • μX = n * P
    • = 8 * 0.25
    • = 2
• The formula to find the standard deviation would be:
  • σ_X = √ n * p * (1 – P)
    • = √ 8 * 0.25 * (1 – 0.25)
    • = √ 8 * 0.25 * (0.75)
    • = √ 1.5
    • = ~1.22
• The hardest part about these questions is recognizing them for what they are. After figuring out that the question is a Bernoulli question, I can then use these simple formulas to find the right answer.

2) Binomial VS Geometric Variables

• The only difference between the two is that a binomial variable has a set number of trials whereas a geometric variable will have continuous trials until a specific outcome occurs.
• The most difficult part about these questions is determining if each trial is independent. I got the one of these questions wrong on my last attempt at the unit test because the question stated that “a different student was picked each time” meaning the trials were not independent which I didn’t understand. 

	Binomial Variable	Geometric Variable
‘Success’ or ‘Failure’	✓	✓
Independent Trials	✓	✓
Fixed Number of Trials	✓
Same Probability per Trial	✓	✓

3) Normal Distributions

• Looks like a bell, which is why it’s also called a ‘bell curve’.
• Uses what’s called standard deviations as a unit of measurement to indicate how the spread of the distribution moves away from the mean.
• Uses the 68-95-99.7 rule which means 68% of the spread of the distribution is +/- 1 S.D. away from the mean, 95% is +/-2 S.D.’s away, and 99.7% is +/-3 S.D.’s away.
• You can combine two normal distributions by simply adding or subtracting their means from each other. When adding two normal distributions to each other, you can also add their respective variances to solve for the new normal distributions’ S.D. When subtracting two normal distributions from each other, however, you still add their variances to determine the new S.D. I think this has something to do with the fact that a S.D. can be positive or negative and therefore you look could potentially be taking two negative S.D.’s when combining normal distributions meaning you would need to take the absolute value of the combined negative values.
• It’s important to remember the z-score formula for these types of questions which is Z_x = (x – μ)/ σ.

4) Expected Value

• Formula for each E(V) is Sum of: (probability of certain event * outcome) = E(V).
• Making tables to organize each event and its respective expected value makes things much more simple. The columns I write down in order are Event, Probability, Outcome, E(V).
• In many questions involving E(V) I’m asked to calculate the probability of the events myself. This often requires me to use combinatorics. This often takes me a few minutes to figure out the proper probability for each specific event.
• I make mistakes on these questions often because I don’t properly comprehend all the details of the question. For instance, some questions state something like “is it a good idea for Johnny to buy the $150 bike light to save on insurance” and I will forget to subtract the $150 from each event.

5) Relative Frequency

• I had one question come up on Relative Frequency which I ended up getting wrong. I was especially annoyed at this since I don’t remember going through videos in this unit about relative frequency. In any case, for the sake of hopefully remembering how to answer it the next time it comers around, the question was:

• Q. Bella has a bunch of toy sets with a different number of toys per set. The number of sets are labeled in the table below except for the number of sets with 3 toys each. She knows that there is an average of 5 pieces per set. How many sets with 3 toys each does she have?

Pieces in each set	Sets	Relative Frequency	Rel. Freq. Pieces/Set
3	x
5	5
10	2
TOTALS:		1.00	5

• The total relative frequency of pieces per set equals 1.00 which I think of as being 100% of the total number of sets (i.e. x + 5 + 2 = 100%).
• The first thing you must do is write out each type of sets relative frequency:

Pieces in each set	Sets	Relative Frequency	Rel. Freq. Pieces/Set
3	x	x/(x + 7)
5	5	5/(x + 7)
10	2	2/(x + 7)
TOTALS:		1.00	5

• Having written in the relative frequency of each set and using x as a variable, you can then multiply the type of sets’ relative frequency by the number of pieces in each type of set to get their corresponding relative frequency of pieces per set:

Pieces in each set	Sets	Relative Frequency	Rel. Freq. * Pieces/Set
3	x	x/(x + 7)	3 * x/(x + 7)
5	5	5/(x + 7)	5 * 5/(x + 7)
10	2	2/(x + 7)	10 * 2/(x + 7)
TOTALS:		1.00	5

• From there you can do some algebra to solve for x:
  • 5 = [3x/(x + 7)] + [25/(x + 7)] + [20/(x + 7)]
  • 5 = (3x + 45)/x + 7
  • 5(x + 7) = 3x + 45
  • 5x + 35 = 3x + 45
  • 5x = 3x + 10
  • 2x = 10
  • x = 5

Pieces in each set	Sets	Relative Frequency	Rel. Freq. Pieces/Set
3	x = 5	x/(x + 7)	3x/(x + 7)
5	5	5/(x + 7)	25/(x + 7)
10	2	2/(x + 7)	20/(x + 7)
TOTALS:	12	1.00	5

Although I didn’t get through the test, I’m without doubt getting much better at probability weighted questions, solving expected values, and solving for the mean and standard deviation of normal distributions. It’s frustrating it’s taking me so long but, as I’ve said before, I think it’s better to be going slowly and really understand the concepts as opposed to rushing through it all without fully grasping the content. I scored 19/21 on my last attempt at the unit test and could should have gotten everything right. I don’t think it will take me too many more attempts to get through it. It would be great if I could get through it early in Week 55 and then get started on the next unit Sampling Distributions (0/700 M.P.) and maybe even finish that unit.

As a final note, this past week I helped run a back-to-school camp at my club which went really well! I worked with 5-9 year old kids on math and a bit of reading/writing. The math was understandably very basic arithmetic but it was still great to actually put some of what I’ve learned to use teaching the kids. I really enjoyed it and it felt like a good step in the right direction for me. It was good experience and also a chance for the manager of that department of the club to see firsthand that I’m fairly good with the kids. Hopefully it helps lead me tutoring more regularly!