Math 1372 – Statistics with Probability – Fall 2014

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  • Exam #2
  • November 19, 2014 at 1:02 pm #18612

    Suman Ganguli
    Participant

    Here is a guide to the questions on the exam:

    #1: rules of probability (Sections 4.3 & 4.5, in particular the Addition Rule and Multiplication Rule)

    #2 & #4: discrete random variables, probability distributions, expected value, variance/standard deviation (Sections 4.2-4.4)

    #3 & #6: binomial random variables (Sec 4.5)

    #6: normal random variables (Sec 6.3)

    If you get stuck, look at the examples and homework exercises for the corresponding sections. You can ask me questions about examples in the book or homework exercises.

    • This topic was modified 9 years, 5 months ago by Suman Ganguli.
    November 19, 2014 at 1:46 pm #18618

    Suman Ganguli
    Participant

    You might find looking at my Google spreadsheet useful for the exam. I’ve included solutions for some of the homework exercises.

    In particular, the “HW7 – Binomial” tab has solutions for a couple of HW exercises from Sec 4.5, including using =binomdist(i,n,p,true) to calculate cumulative binomial probabilities–which could be useful for #3 & #6 on the exam

    You can view the Google spreadsheet here: https://docs.google.com/spreadsheets/d/1pgVX90vBiFrJgdiTKfqIv23Kv_eJC_v0lwD89upj2Rk/edit?usp=sharing

    November 21, 2014 at 8:59 pm #18654

    Janatul Mawa
    Participant

    Hello Professor, i’m confused in q#2(a). there is only one blue marble, if two marble drawn without replacement then there won’t be any marble left for second time..i was wondering if Y would be 0. cause it would be like (1/10)X (0/9) which will give us 0… i’m just confused if i’m getting it right..

    November 22, 2014 at 12:18 pm #18665

    Suman Ganguli
    Participant

    You’re on the right track. The calculation you did shows that the probability of drawing two blue marbles is 0–that makes sense since as you said there’s only one blue marble to start, so it’s impossible to draw two. But that doesn’t mean the only possible value of Y is 0. It shows that 2 is not a possible value of Y.

    Think about the other possibilities for you might draw (hint: look at part (b)), and consider what the value of Y is in each of those cases.

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