Math 1372 – Statistics with Probability – Fall 2014

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  • #26638

    Suman Ganguli
    Participant

    Review the following topics/exercises (note that solutions for all of these exercises are in my Google spreadsheet):

    (1) Discrete random variables and their probability distributions

    Exam #2: #2

    HW #6: Sec 5.3 #11, 14, 19; Sec 5.4 #14 (start by writing down probability distribution)

    (2) Continuous random variables

    HW #7: Sec 6.2 #1,3

    (3) Normal random variables

    Exam #2: #5
    HW #8: Ch 6 Review exercises

    (4) Sample Means and Central Limit Theorem

    HW #9:
    Sec 7.3: #5, 6
    Sec 7.4: #1, 13, 15

    #28381

    Suman Ganguli
    Participant

    I received a question via email about Exercises #5 & #6 from Sec 7.3. I thought I’d post my reply, in case it might help you understand these questions and concepts:

    These exercises are applications of the Central Limit Theorem, as we stated it in class (it’s stated slightly differently in Sec 7.4 of the book). Namely, if you take samples of size n from a population with mean mu and standard deviation sigma, then the sample means X-bar are normally distributed with the same mean mu but with standard deviation equal to sigma/square root(n).

    In Sec 7.1 of the book, these facts about the mean and standard deviation of X-bar (the sample means) are on p300, but the latter is stated in terms of the variance instead of standard deviation:

    E[X-bar] = mu
    Var(X-bar) = sigma^2/n

    If you take the square root of both sides of the latter equation, you get how we stated it in class, in terms of the standard deviation of sample means:
    SD(X-bar) = sigma/square root(n)

    For exercises 5 & 6, you are given the population mean mu and standard deviation sigma, and asked to calculate E[X-bar] and SD(X-bar) for different sample sizes n. But E[X-bar] = mu in every case, so you just need to calculate SD(X-bar) = sigma/sqrt(n) for each n.

    In #5, you are given that mu = 2.4 mg (mean nicotine content) with sigma = 0.2 (std dev). In every case, the expected value of the sample mean will be the same as the population mean, i.e.,

    E[X-bar] = 2.4

    For part (a), if you take a sample of n = 36 randomly selected cigarettes, the standard deviation of sample means will be

    SD(X-bar) = 0.2/sqrt(36) = 0.2/6 = 0.0333

    Actually #5 asks you to calculate variance, but that’s just the standard deviation squared:

    Var(X-bar) = sigma^2/n = (0.2)^2/36 = 0.04/36 = 0.001111

    For parts (b)-(d), you just need to change 36 in the standard deviation/variance calculations to 64, 100, and 900 respectively.

    I hope that helps!

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