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Math Help - Probability that celebrities are less likely to die in the month of their birthday

  1. #1
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    Probability that celebrities are less likely to die in the month of their birthday

    I am not sure whether I have approached this question correctly... I would appreciate your input on my answer and any suggestions regarding any other calculations I should be doing. Thanks in advance.

    Question:

    Evidence has been produced that famous people are less likely to die in the month of their birthday than in other months. The (skeptical) hypothesis is that dying is equally likely in any month regardless of birthday. Suppose that out of 120 celebrities 7 died in the month of their birthday. Imagine a hat with 12 cards, each card a month, as well as a list of the 120 celebrity birthdays. We shuffle and pick a card, noting whether it matched the first celebrity birth month. We then repeat this (replacing the card each time, of course), each time noting whether the month picked from the hat matched the next birth month, etc., until we have gone all the way through the 120 names on the list.

    Then we repeat this procedure 100 times, each time recording how many matches we got between the 120 picks from the hat, and the list of 120 birthdays. We got the following frequency distribution. What is your conclusion and why?

    The column on the left represents the number of celebrities dying in their birthday month (6, 7, 8 etc). The column on the right represents the frequency. (1, 3, 9 etc). (I cannot paste the table properly here - if there is a way of doing this, let me know).

    6 1
    7 3
    8 9
    9 20
    10 32
    11 25
    12 7
    13 1
    14 2

    My answer so far:

    The probability of observing 7 or fewer matches (celebrities dying in their birth month) is 4/100 = 1/25.



    I do not know if this is correct or enough to answer this question sufficiently.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Well, I'm not sure, but I would find the expected number of deaths on their birthday.

    I get 9.99 = 10

    So, I would conclude that out of 120 celebrities, only 10 die on their birthday, which is also a probability of 1/12, which means I would support the skeptical hypothesis, that is dying is equally likely in any month regardless of birthday.
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    Well, I'm not sure, but I would find the expected number of deaths on their birthday.

    I get 9.99 = 10

    So, I would conclude that out of 120 celebrities, only 10 die on their birthday, which is also a probability of 1/12, which means I would support the skeptical hypothesis, that is dying is equally likely in any month regardless of birthday.
    Thanks for this...how did you work out 9.99 = 10?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Oh, it's by substituting the frequency by the probabilities.

    I'll display it as a horizontal table.

    \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline<br />
x&6&7&8&9&10&11&12&13&14 \\ \hline<br />
f&1&3&9&20&32&25&7&1&2 \\ \hline<br />
P(x)&0.01&0.03&0.09&0.2&0.32&0.25&0.07&0.01&0.02 \\ \hline<br />
\end{array}

    Then, the expectation is given by:

    \displaystyle \Sigma (xP(x)) = (6\times 0.01) + (7\times 0.03) + (8\times 0.09) + ... + (14 \times 0.02) = 9.99
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  5. #5
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    Quote Originally Posted by Unknown008 View Post
    Oh, it's by substituting the frequency by the probabilities.

    I'll display it as a horizontal table.

    \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline<br />
x&6&7&8&9&10&11&12&13&14 \\ \hline<br />
f&1&3&9&20&32&25&7&1&2 \\ \hline<br />
P(x)&0.01&0.03&0.09&0.2&0.32&0.25&0.07&0.01&0.02 \\ \hline<br />
\end{array}

    Then, the expectation is given by:

    \displaystyle \Sigma (xP(x)) = (6\times 0.01) + (7\times 0.03) + (8\times 0.09) + ... + (14 \times 0.02) = 9.99
    Thanks very much. By the way if you or anyone could explain how to paste tables on this forum that would be much appreciated.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    You can make use of the 'math' tabs.

    Then, double click on my table to see the code. The tables took me a while to know how to use it properly though.

    Or, here it is:

    [math ]\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline
    x & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline
    f & 1 & 3 & 9 & 20 & 32 & 25 & 7 & 1 & 2 \\ \hline
    P(x) & 0.01 & 0.03 & 0.09 & 0.2 & 0.32 & 0.25 & 0.07 & 0.01 & 0.02 \\ \hline
    \end{array}[/tex]

    I forgot how to post the tabs without activating it... so, in this example, remove the space after math.

    each c represents a column.
    '|' this symbol is a separation. It is not necessary.
    & is to switch column
    \\ is to switch row
    \hline is a command to insert a horizontal line as separation. This too isn't necessary but I use them for aesthetic purposes
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  7. #7
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    Quote Originally Posted by Unknown008 View Post
    You can make use of the 'math' tabs.

    Then, double click on my table to see the code. The tables took me a while to know how to use it properly though.

    Or, here it is:

    [math ]\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline
    x & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline
    f & 1 & 3 & 9 & 20 & 32 & 25 & 7 & 1 & 2 \\ \hline
    P(x) & 0.01 & 0.03 & 0.09 & 0.2 & 0.32 & 0.25 & 0.07 & 0.01 & 0.02 \\ \hline
    \end{array}[/tex]

    I forgot how to post the tabs without activating it... so, in this example, remove the space after math.

    each c represents a column.
    '|' this symbol is a separation. It is not necessary.
    & is to switch column
    \\ is to switch row
    \hline is a command to insert a horizontal line as separation. This too isn't necessary but I use them for aesthetic purposes
    Thanks - I'll try this.
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  8. #8
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    Quote Originally Posted by Unknown008 View Post
    Oh, it's by substituting the frequency by the probabilities.

    I'll display it as a horizontal table.

    \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline<br />
x&6&7&8&9&10&11&12&13&14 \\ \hline<br />
f&1&3&9&20&32&25&7&1&2 \\ \hline<br />
P(x)&0.01&0.03&0.09&0.2&0.32&0.25&0.07&0.01&0.02 \\ \hline<br />
\end{array}

    Then, the expectation is given by:

    \displaystyle \Sigma (xP(x)) = (6\times 0.01) + (7\times 0.03) + (8\times 0.09) + ... + (14 \times 0.02) = 9.99
    There was just one other thing - can you define the \displaystyle \Sigma (xP(x)) in simple English?.... I not not familiar with the mathematical symbols.
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  9. #9
    MHF Contributor Unknown008's Avatar
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    \Sigma (xP(x))

    \Sigma is the total sum, denoted by capital sigma.

    \Sigma (xP(x)) is the total sum of the product of x and P(x)

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