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Statistics QUESTION CIE AS LEVEL ; help pleased watch

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    b) A normal distribution has mean μ and standard deviation σ. If 800 observations are taken from
    this distribution, how many would you expect to be between μ − σ and μ + σ? [3]


    Can someone please explain this to me . Thank you
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    (Original post by Sammysammy99)
    b) A normal distribution has mean μ and standard deviation σ. If 800 observations are taken from this distribution, how many would you expect to be between μ − σ and μ + σ? [3]


    Can someone please explain this to me . Thank you
    You need to refer to:

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    I'd calculate what proportion of the data lies between μ − σ and μ + σ on a normal distribution graph(i.e. proportion of data between -1 and 1). I calculate it to be 0.682689

    Then multiply this by 800 to get 546.152. So I'd expect approximately 546 of these observations to lie within these parameters.

    Any steps of that you don't understand why I've done it?
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    (Original post by AzureCeleste)
    I'd calculate what proportion of the data lies between μ − σ and μ + σ on a normal distribution graph(i.e. proportion of data between -1 and 1). I calculate it to be 0.682689

    Then multiply this by 800 to get 546.152. So I'd expect approximately 546 of these observations to lie within these parameters.

    Any steps of that you don't understand why I've done it?
    I don't get where the number 0.682689 came from I'm really confused?
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    (Original post by Sammysammy99)
    I don't get where the number 0.682689 came from I'm really confused?
    I used my calculator to get it by doing normcdf(-1,1,0,1). I don't know however if you normally use a calculator for this?
    If you don't, you can refer to the diagram posted further up to get 68.2% (aka 0.682)
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    (Original post by Sammysammy99)
    I don't get where the number 0.682689 came from I'm really confused?
    How do u know the middle part has a 34% ?
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    (Original post by Sammysammy99)
    How do u know the middle part has a 34% ?
    Look at the graph RDKGames posted.
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    (Original post by AzureCeleste)
    Look at the graph RDKGames posted.
    I get that graph but why did he put 34 percent in there ? From where did he get those numbers
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    (Original post by Sammysammy99)
    I get that graph but why did he put 34 percent in there ? From where did he get those numbers
    Probably got it off the internet???
    What kind of calculations have you done with the normal graph?
    Like do you know how to calculate the proportion of data lying between 0 and and 1 σ ?
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    (Original post by Sammysammy99)
    How do u know the middle part has a 34% ?
    This is something that should be covered in your lessons.

    Take X \sim N(0,1) which means X is your uniform random variable following a normal distribution.

    The mean here is \mu = 0, and the variance here is \sigma^2 = 1 therefore \sigma = \sqrt{1} = 1 which is also our standard deviation.

    To find the percentage of values which lie between our mean and one standard deviation ahead of it, we do:
    \begin{aligned} P(\mu < X < \mu + \sigma) & = P(0 < X < 1) \\ & = P(X<1)-P(X<0) \\ & = P(X<1)-0.5

    Now we get that P(X<1) \approx 0.8413 from the normal distribution table.
    Therefore we have P(0 < X < 1) \approx 0.3413 which is 34.1% and hence that's where it comes from on the image I posted above.

    If you cannot see the proper mathematical form in this post, I suggest you get off your app and use your phone's browser to view maths on TSR for it to display properly.
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    (Original post by RDKGames)
    This is something that should be covered in your lessons.

    Take X \sim N(0,1) which means X is your uniform random variable following a normal distribution.

    The mean here is \mu = 0, and the variance here is \sigma^2 = 1 therefore \sigma = \sqrt{1} = 1 which is also our standard deviation.

    To find the percentage of values which lie between our mean and one standard deviation ahead of it, we do:
    \begin{aligned} P(\mu < X < \mu + \sigma) & = P(0 < X < 1) \\ & = P(X<1)-P(X<0) \\ & = P(X<1)-0.5

    Now we get that P(X<1) \approx 0.8413 from the normal distribution table.
    Therefore we have P(0 < X < 1) \approx 0.3413 which is 34.1% and hence that's where it comes from on the image I posted above.

    If you cannot see the proper mathematical form in this post, I suggest you get off your app and use your phone's browser to view maths on TSR for it to display properly.
    Ok thank you very much for this detailed answer , really appreciate it , I just do not get why u didn’t add one the the left side , what I mean is that u added r to the right side e.g mu+s.d but only to the right side
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    (Original post by Sammysammy99)
    Ok thank you very much for this detailed answer , really appreciate it , I just do not get why u didn’t add one the the left side , what I mean is that u added r to the right side e.g mu+s.d but only to the right side
    Not sure what you mean?

    Are you referring to \mu < X < \mu + \sigma? Why would I add one here...??

    As I mentioned early in the post, \mu = 0 and \sigma = 1, so the region \mu < X < \mu + \sigma is the same as  0 < X < 0+1
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    (Original post by RDKGames)
    Not sure what you mean?

    Are you referring to \mu < X < \mu + \sigma? Why would I add one here...??

    As I mentioned early in the post, \mu = 0 and \sigma = 1, so the region \mu < X < \mu + \sigma is the same as  0 < X < 0+1
    No it’s fine , thank you veryyy much I get it now , I went over it and understood . Very helpful thank u
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