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    I don't understand how to answer the following question:

    In a population of men the systolic blood pressure shows a normal distribution. The mean of the population is 125 and (measured in mm Hg) and the standard deviation is 10. If the population was 1000, how many of them have a blood pressure between 115 and 135mm Hg?

    The AQA text book says the answer is 680 but I do not understand how to get there.

    Any help?

    Cheers.
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    (Original post by AshleyWheat)
    I don't understand how to answer the following question:

    In a population of men the systolic blood pressure shows a normal distribution. The mean of the population is 125 and (measured in mm Hg) and the standard deviation is 10. If the population was 1000, how many of them have a blood pressure between 115 and 135mm Hg?

    The AQA text book says the answer is 680 but I do not understand how to get there.

    Any help?

    Cheers.
    Firstly, if its a normal distribution, then you would model it like this.

    X ~ N(125,100) Where 125 is the mean and 100 is the variance(SD is the square root of the variance).

    you are trying to find the number of people between 115 and 135, so the probability will be
    p(115<X<135)
    To compare this to the standard normal, you use the equation Z = (X-mean)/standard deviation
    this gives you p((115-125)/10<Z<(135-125)/10)
    I get this as p(-1<Z<1)
    once you have this, you have to look in the tables, I can't get latex of the 'fi' symbol so I'll use o instead.
    p(-1<Z<1) = p(Z<1) - p(Z<-1)
    = p(Z<1) - (1 - p(Z<1))
    = p(Z<1) + p(Z<1) -1
    =2 x p(Z<1) -1
    That was there for my reference, because I find it easier to do that instead, now in the tables I get p(Z<1) = 0.8413
    2x0.8413 -1 = 0.6826
    So back to your question, if the population was 1000, then you would simply multiply the probability by the population size
    0.6826 x 1000 = 682.6

    the question may have said to 2 significant figures, if so this would be 680.

    I don't think this is a Biology question, I covered this in statistics 2.
 
 
 
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