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Math Help - Estimation,Confidence intervals and tests

  1. #1
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    Cool Estimation,Confidence intervals and tests

    Can anyone help me to solve the following problem?

    The 25 children in a class each roll a fair die 30 times and record the number of sixes they obtain. Find an estimate of the probability that the mean number of sixes recorded for the class is less than 4.5
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    Quote Originally Posted by Sashikala View Post
    Can anyone help me to solve the following problem?

    The 25 children in a class each roll a fair die 30 times and record the number of sixes they obtain. Find an estimate of the probability that the mean number of sixes recorded for the class is less than 4.5
    The expected value or mean number of 6's rolled for each child will be 30/6 = 5 and as there are 25 students the mean number of 6's rolled for the calss will be 750/6 = 125.

    Is there any indication of how the data is spread? I.e if the data is normal and has a standard deviation then you can use a z-score

    For example (x<4.5) = (Z< (4.5 - mean)/std)

    Then we can find a value in a standard normal distribution table.

    I have created a simulation for you below, could be helpful!
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  3. #3
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    Cool Estimation,Confidence intervals and tests

    Thanks for the response.
    Yes, We need to use the following
    P(x<4.5) = P(Z< (4.5 - mean)/std)

    Here mean will be 25*(30/6)/25=5
    How to get the value for std?
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  4. #4
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    Quote Originally Posted by Sashikala View Post
    Thanks for the response.
    Yes, We need to use the following
    P(x<4.5) = P(Z< (4.5 - mean)/std)

    Here mean will be 25*(30/6)/25=5
    How to get the value for std?
    I would say that because we are looking for the probabilty of throwing a 6 then the problem can be looked as a binomial random variable.

    For the entire class we have
    n = 750,
    p (success) = 1/6,
    q = 1-p (failure) = 5/6

    the standard deviation \sigma can be found as

    \sigma = \sqrt{npq}

    \sigma = \sqrt{750.\frac{1}{6}.\frac{5}{6}}

    \sigma = 10.21

    Or per student

    n = 30,
    p (success) = 1/6,
    q = 1-p (failure) = 5/6

    the standard deviation \sigma can be found as

    \sigma = \sqrt{npq}

    \sigma = \sqrt{30.\frac{1}{6}.\frac{5}{6}}

    \sigma = 2.04
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  5. #5
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    Talking Estimation,Confidence intervals and tests

    You explained it very clearly.
    Oh...., Thank you very much.!!!
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