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Math Help - Probability: health insurance

  1. #1
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    Probability: health insurance

    It is believed that 20% of Americans do not have any health insurance. Suppose this is true and let X equal the number with no health insurance in a random sample of 15 Americans.

    How is X distributed? I think its b(15,.20)

    What is the mean, variance and standard deviation of X?

    What is P(X \ge5)?
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  2. #2
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    Quote Originally Posted by calabrone View Post
    It is believed that 20% of Americans do not have any health insurance. Suppose this is true and let X equal the number with no health insurance in a random sample of 15 Americans.

    How is X distributed? I think its b(15,.20) Mr F says: Correct.

    What is the mean, variance and standard deviation of X?

    Mr F says: Substitute the values of n and p into the formulae you should have been given.

    What is P(X \ge5)?
    \Pr(X \geq 5) = 1 - \Pr(X \leq 4)  = 1 - [\Pr(X = 4) + \Pr(X = 3) + \Pr(X = 2) + \Pr(X = 1) + \Pr(X = 0)].

    Either use the formula to calculate each probability or use technology.

    All formula can be found here if, for example, your dog ate the page of your notes that had them: Binomial distribution - Wikipedia, the free encyclopedia
    Last edited by mr fantastic; April 10th 2009 at 08:59 PM. Reason: Fixed some latex (I was waiting for Halls to fix it for me lol!).
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    I am having such a hard time because I have no notes. Would the mean be 15 * .20 = 3? and variance be (15*0.20)(1-0.20)=2.4?

    I'm not sure how to calculate the standard deviation and P(X \ge5).
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    Quote Originally Posted by calabrone View Post
    I am having such a hard time because I have no notes. Would the mean be 15 * .20 = 3? and variance be (15*0.20)(1-0.20)=2.4? Mr F says: Yes. And {\color{red}\text{sd} = \sqrt{\text{Variance}}}.

    I'm not sure how to calculate the standard deviation and P(X \ge5).
    In my previous post (I've fixed the latex) you will see that there are a number of probabilities you need to calculate. Note that \Pr(X = r) = ^{15}C_r (0.2)^r (1 - 0.2)^{15 - r}. Substitute r = 0, 1, 2, 3, 4 and calculate the required probabilities.
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  5. #5
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    so SD= radical 2.4

    for P(X \ge5):

    1-[P(X=4)+P(X=3)+P(X=2)+P(X=1)+P(X=0)]=1-(4*1/5+3*1/5+2*1/5+1*1/5+0*1/5)=1-(4/5+3/5+2/5+1/5+0)=1

    My answer doesn't seem correct. I feel like I should have included the probability of HAVING health insurance for the remaining.... for example, for X=4 there is an 80% chance for each of the the remaining 11 (15-4) to have insurance.
    Last edited by calabrone; April 11th 2009 at 09:37 AM.
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    Quote Originally Posted by calabrone View Post
    so SD= radical 2.4 Mr F says: Yes.

    for P(X \ge5):

    1-[P(X=4)+P(X=3)+P(X=2)+P(X=1)+P(X=0)]=1-(4*1/5+3*1/5+2*1/5+1*1/5+0*1/5)=1-(4/5+3/5+2/5+1/5+0)=1

    My answer doesn't seem correct. I feel like I should have included the probability of HAVING health insurance for the remaining.... for example, for X=4 there is an 80% chance for each of the the remaining 11 (15-4) to have insurance.
    You have not correctly used the formula I gave you for calculating the probabilities. Please show all your working for how you calculated Pr(X = 1), say, so that what you're doing wrong can be pointed out.

    By the way: You should realise that 4/5+3/5+2/5+1/5+0 adds up to more than 1, which is clearly absurd. This should have told you that your numerical calculations were wrong.
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    1-[(4/15 * 1/20) + (3/15 * 1/20) + (2/15 * 1/20) + (1/15 * 1/20) + (0/15 * 1/20)]=1-10/300=290/300=29/30? This answer doesn't make sense either. I don't know why I'm so confused.
    Last edited by calabrone; April 11th 2009 at 03:38 PM.
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  8. #8
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    Quote Originally Posted by calabrone View Post
    1-[(4/15 * 1/20) + (3/15 * 1/20) + (2/15 * 1/20) + (1/15 * 1/20) + (0/15 * 1/20)]=1-10/300=290/300=29/30? This answer doesn't make sense either. I don't know why I'm so confused.
    Quote Originally Posted by mr fantastic View Post
    [snip]
    Please show all your working for how you calculated Pr(X = 1), say, so that what you're doing wrong can be pointed out.

    [snip]
    If you did what I asked then progress might be made. Show how you used the formula I gave you to calculate Pr(X = 1).
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  9. #9
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     <br />
P(x=1)=15!/1!14!(0.2)^{1}(1-0.2)^{14}=15*0.2*0.04398=0.13194<br />
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  10. #10
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    Quote Originally Posted by calabrone View Post
     <br />
P(x=1)=15!/1!14!(0.2)^{1}(1-0.2)^{14}=15*0.2*0.04398=0.13194<br />
    Correct. Now calculate the other probabilities. Then get the final answer (which is 0.1642, correct to 4dp).
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  11. #11
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    Thank you so much! I was beginning to feel hopeless. I don't know how I'm going to make it through this class when this is only the beginning. Thank you for taking so much time to help me! Hopefully my calculations give me the same answer.
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