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Math Help - Normal Distributions

  1. #1
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    Normal Distributions

    all are normal distributions

    1) Sample size = 6. mean of 10 and variance of 2. Calculate the probability that 2 of the values are less than 9 and the other 4 are greater than nine.

    I can find the probability of 1 being greater than or less than 9. That would just be (8-10)/(sqrt 2) right?

    2)The sizes of claims are normal random variables. Mean = 1800. SD=400. Find probability that 2 random claims differ by more than 500.



    3) n = 80 they are independent. indiv mean = 574 with sd = 186. P[average claim <565]

    P[Z<565-574/(Sqrt(186^2/80)). that gave me -.433 as my Z and then i took 1-.6664 (chart value of .433). I got .3336 as my answer.
    I got (sqrt(186^2/80) as my variance for each average term. the math was 1/80^2 * var(s)= 80*Var/(80^2). Var=SD^2 so that gave me (sqrt(186^2/80)

    4) X = rate of interest. mean=.06 SD=.015 Amount in account after year => Y=10,000 e^x Find E[x] after a year
    Last edited by PensFan10; February 4th 2009 at 10:53 AM.
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    Quote Originally Posted by PensFan10 View Post
    all are normal distributions

    1) Sample size = 6. mean of 10 and variance of 2. Calculate the probability that 2 of the values are less than 9 and the other 4 are greater than nine.

    I can find the probability of 1 being greater than or less than 9. That would just be (8-10)/(sqrt 2) right? Mr F says: Can a probabilty be negative, I wonder .....? Surely you mean the z-value, in which case it is actually (9 - 10)/(sqrt 2) = -0.7071.

    [snip]
    I assume you mean that the sample is taken from a normal population whose mean is 10 and variance is 2 ....

    Pr(Z < -0.7071) = 0.23975 (correct to 5 decimal places)

    Now let Y be the random variable number of values less than 9.

    Y ~ Binomial(n = 6, p = 0.23975).

    Calculate Pr(Y = 2).
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    Quote Originally Posted by PensFan10 View Post
    [snip]
    2)The sizes of claims are normal random variables. Mean = 1800. SD=400. Find probability that 2 random claims differ by more than 500.

    [snip]
    Apply a well known result for the difference of two independent normal random variables:

    U = X_1 - X_2 ~ N(\mu = 1800 - 1800 = 0, \sigma^2 = 400^2 + 400^2 = 3200).

    Calculate \Pr(U \geq 500).
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    Is the probability Y=2 (0.23975)(0.23975) = .05748?

    Then for Greater than nine:
    probability .500
    then for the other 4 (.500)^4 = .0625?
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    Quote Originally Posted by PensFan10 View Post
    [snip]4) X = rate of interest. mean=.06 SD=.015 Amount in account after year => Y=10,000 e^x Find E[x] after a year
    E(10,000 e^X) = 10,000 E(e^X).

    E(e^X) = \int_{-\infty}^{+\infty} e^x \, f(x) \, dx

    where f(x) is the pdf of the given normal distribution (the formula will be in your class notes or textbook).

    I assume an approximate answer is sufficient ie. use technology to get a decimal approximation of the integral.

    (I cannot understand your question 3. PLease post the whole question, exactly as it's written in your textbook or wherever it is you got it from).
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    for number 2

    500-0/(3200) = .1563

    .1563 = .5636
    thats for less than.
    greater than would be 1-(.5636)= .4364
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    number 3 =
    An insurance portfolio consists of 80 policies. The claim amount for the policies are independent, identically distributed random variables with a normal distribution. The individual mean claim is 574 with a SD of 186. Find the probability that the average claim is below 565
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    Quote Originally Posted by PensFan10 View Post
    Is the probability Y=2 (0.23975)(0.23975) = .05748?

    Then for Greater than nine:
    probability .500
    then for the other 4 (.500)^4 = .0625?
    No. And I cannot follow where this working has come from (nor will your instructor be able to either, I'll bet).

    Use the binomial distribution in the way suggested by my post. Have you been taught the binomial distribution?

    I get 0.288 (correct to three decimal places).


    And please .... quote the posts that you are asking about (use the Quote tool at the bottom right).
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    Quote Originally Posted by PensFan10 View Post
    number 3 =
    An insurance portfolio consists of 80 policies. The claim amount for the policies are independent, identically distributed random variables with a normal distribution. The individual mean claim is 574 with a SD of 186. Find the probability that the average claim is below 565
    Use the sampling distribution of the mean: Read this thread: http://www.mathhelpforum.com/math-he...stic-help.html
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    Quote Originally Posted by PensFan10 View Post
    500-0/(3200) = .1563

    .1563 = .5636
    thats for less than.
    greater than would be 1-(.5636)= .4364
    This working makes no sense at all. How can 0.1563 = 0.5636? It doesn't matter what you mean, what matters is what you've written. Statements like 0.1563 = 0.5636 are ridiculous. Please post working that is well set out and that makes sense if you expect people to read and comment on your work.

    The question requires you to calculate Pr(U > 0.1563). Solve this in the way you've been taught how to deal with normal probabilities.
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    Quote Originally Posted by mr fantastic View Post
    No. And I cannot follow where this working has come from (nor will your instructor be able to either, I'll bet).

    Use the binomial distribution in the way suggested by my post. Have you been taught the binomial distribution?

    I get 0.288 (correct to three decimal places).


    And please .... quote the posts that you are asking about (use the Quote tool at the bottom right).
    We went over it in one class. We just did some basic example.

    I assume you mean that the sample is taken from a normal population whose mean is 10 and variance is 2 ....

    Pr(Z < -0.7071) = 0.23975 (correct to 5 decimal places)

    Now let Y be the random variable number of values less than 9.

    Y ~ Binomial(n = 6, p = 0.23975).

    Calculate Pr(Y = 2).
    I figured that .23975 was probability that value is less than nine. Then to get the probability that the figure is greater than nine i did this:
    10- 10/2 = which gave me 0. the probability for the z value is .500.

    I used 10-10 because 10 is the first number greater than nine and 10 is the mean.

    The probability that corresponds to 0 = .500. I raised .500 to the fourth because there are 4 numbers selected to be greater than 9.

    (.500)^4 = .0625
    (0.23975)^2 = .05748
    then add them together to get = .11998
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  12. #12
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    Quote Originally Posted by PensFan10 View Post
    We went over it in one class. We just did some basic example.



    I figured that .23975 was probability that value is less than nine. Then to get the probability that the figure is greater than nine i did this:
    10- 10/2 = which gave me 0. the probability for the z value is .500.

    I used 10-10 because 10 is the first number greater than nine and 10 is the mean.

    The probability that corresponds to 0 = .500. I raised .500 to the fourth because there are 4 numbers selected to be greater than 9.

    (.500)^4 = .0625
    (0.23975)^2 = .05748
    then add them together to get = .11998
    Do it as I outlined in post #2.
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    Quote Originally Posted by mr fantastic View Post
    Apply a well known result for the difference of two independent normal random variables:

    U = X_1 - X_2 ~ N(\mu = 1800 - 1800 = 0, \sigma^2 = 400^2 + 400^2 = 3200).

    Calculate \Pr(U \geq 500).
    Mean = 0 with the variance = 3200

    What i did was take the X-Mean/(variance)
    500-0/3200 = .15625 as my z value

    From this i got 500-0/3200
    P[Z(greater than or equal to)500] = since the z value of .16 is .5636
    since the z value gives numbers less than i subtracted .5636 from 1 which gave me:
    1-.5636 = .4364
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    Quote Originally Posted by mr fantastic View Post
    I assume you mean that the sample is taken from a normal population whose mean is 10 and variance is 2 ....

    Pr(Z < -0.7071) = 0.23975 (correct to 5 decimal places)

    Now let Y be the random variable number of values less than 9.

    Y ~ Binomial(n = 6, p = 0.23975).

    Calculate Pr(Y = 2).
    Okay, I see how you got 0.23975.
    Using the Binomial theorem I get
    (6C2)(.23975)^2(.76025)^4 = .2880
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  15. #15
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    Quote Originally Posted by PensFan10 View Post
    Mean = 0 with the variance = 3200

    What i did was take the X-Mean/(variance)
    500-0/3200 = .15625 as my z value Mr F says: 3200 is NOT the sd .... It's the variance. Your z-value is therefore vey wrong.

    From this i got 500-0/3200
    P[Z(greater than or equal to)500] = since the z value of .16 is .5636
    since the z value gives numbers less than i subtracted .5636 from 1 which gave me:
    1-.5636 = .4364
    Please take greater care.
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