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    Please could anyone help me on part 2aiii, 6b and 6d

    On 2aiii I get that they've done cirumfurence= pi x diameter
    which gives

    pi ( X + 10/pi )

    This is given on the mark scheme
    But then they multiply E(X) by pi and add 10 .. why do you add on 10 and not 10/pi ?? Does the denominator have an effect on the mean?

    And then 6b, I understand that F(m) = 1/2
    But I just don't know what to do from there
    On the MS they seem to have just subbed 1 in? But I don't get it, I set F(x) equal to 1/2 and was left with a cubic that I couldn't do anything with

    For 6c, I don't understand how they've found P(X<1.5), and just don't really understand this whole question part!

    Thanks so so much for any help, and if you have any good statistics resources please let me know as I'm struggling with some papers! Thanks again
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    (Original post by jazz_xox_)
    Please could anyone help me on part 2aiii, 6b and 6d

    On 2aiii I get that they've done cirumfurence= pi x diameter
    which gives

    pi ( X + 10/pi )

    This is given on the mark scheme
    But then they multiply E(X) by pi and add 10 .. why do you add on 10 and not 10/pi ?? Does the denominator have an effect on the mean?
    \mathbb{E}[aX+b] = a \mathbb{E}[X]+b and we have C = \pi X + 10 so there's that.

    And then 6b, I understand that F(m) = 1/2
    But I just don't know what to do from there
    On the MS they seem to have just subbed 1 in? But I don't get it, I set F(x) equal to 1/2 and was left with a cubic that I couldn't do anything with
    When it says 'verify' all you need to do is subsitute in the value they say and see whether the statement holds. This is what they did, just sub in 1 and get 1/2. Job done.

    For 6c, I don't understand how they've found P(X<1.5), and just don't really understand this whole question part!
    I assume you're referring to 6d... in which case P(X&lt;1.5) = \int_0^{1.5} f(x) .dx which requires two different integrals since f is made up of two different functions over this region. Alternatively, you can determine P(X&gt;1.5) and then just subtract this from 1. This is given by \displaystyle P(X&gt;1.5) = \int_{1.5}^2 f(x) .dx = \int_{1.5}^2 \frac{1}{4}(5-2x).dx
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    (Original post by RDKGames)
    \mathbb{E}[aX+b] = a \mathbb{E}[X]+b and we have C = \pi X + 10 so there's that.



    When it says 'verify' all you need to do is subsitute in the value they say and see whether the statement holds. This is what they did, just sub in 1 and get 1/2. Job done.



    I assume you're referring to 6d... in which case P(X&lt;1.5) = \int_0^{1.5} f(x) .dx which requires two different integrals since f is made up of two different functions over this region. Alternatively, you can determine P(X&gt;1.5) and then just subtract this from 1. This is given by \displaystyle P(X&gt;1.5) = \int_{1.5}^2 f(x) .dx = \int_{1.5}^2 \frac{1}{4}(5-2x).dx
    Thank you so so much!! Can I ask what resources you used when you did a-level maths (presuming you're above that level?)

    Also this question may make no sense at all, but I'll try
    When finding a probability e.g. P(2 ≤ X ≤ 5)
    You'd usually do P(X ≤ 5) - P(X ≤ 2)

    But I've noticed that in continuous random variable questions, you'd just do F(5) - F(2) i.e. the less than or equal to symbol has no effect. Is this for continuous questions only? So never for discrete, poisson etc?
    Thanks again
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    (Original post by jazz_xox_)
    Thank you so so much!! Can I ask what resources you used when you did a-level maths (presuming you're above that level?)
    I only have my uni resources for this stuff as I didn't cover it at A-Level.

    Also this question may make no sense at all, but I'll try
    When finding a probability e.g. P(2 ≤ X ≤ 5)
    You'd usually do P(X ≤ 5) - P(X ≤ 2)

    But I've noticed that in continuous random variable questions, you'd just do F(5) - F(2) i.e. the less than or equal to symbol has no effect. Is this for continuous questions only? So never for discrete, poisson etc?
    Thanks again
    Very valid question. Indeed, with a continuous random variable X, get that P(X \leq 1) = P(X &lt; 1). This is because P(X\leq 1) = P(X &lt; 1) + P(X = 1) but we know (and can very trivially prove) that P(X = 1) = 0 due to X being continuous.

    This is not generally applicable to discrete random variables.
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    (Original post by RDKGames)
    I only have my uni resources for this stuff as I didn't cover it at A-Level.



    Very valid question. Indeed, with a continuous random variable X, get that P(X \leq 1) = P(X &lt; 1). This is because P(X\leq 1) = P(X &lt; 1) + P(X = 1) but we know (and can very trivially prove) that P(X = 1) = 0 due to X being continuous.
    This is not applicable to discrete variables.
    Ohh okay fair enough! And that makes sense, thank you!
 
 
 
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