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Stats 2 questions

Please could someone explain questions 6b and c :smile:

I'm unsure on c) why it is set equal to 1/4 if we are using the upper quartile?
And on d), why are we finding P(X>1.5) as part of the working out as oppose to P(X<1.5)?

Thank you :smile:

Paper: http://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2011/june/AQA-MS2B-W-QP-JUN11.PDF

Mark scheme: http://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2011/june/AQA-MS2B-W-MS-JUN11.PDF
Original post by jazz_xox_
Please could someone explain questions 6b and c :smile:

I'm unsure on c) why it is set equal to 1/4 if we are using the upper quartile?
And on d), why are we finding P(X>1.5) as part of the working out as oppose to P(X<1.5)?

Thank you :smile:

Paper: http://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2011/june/AQA-MS2B-W-QP-JUN11.PDF

Mark scheme: http://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2011/june/AQA-MS2B-W-MS-JUN11.PDF


b) This is simply from the fact that the median, mm, is a point along the distribution of XX, FX(x)F_X(x), so that FX(m)=12F_X(m) = \frac{1}{2} (i.e. the middle value of the distribution) - so then just plug in x=1x=1 to obtain F(1)=12F(1)=\frac{1}{2} and this verifies that it's the median.

c) This is because we have ∫0qf(x).dx=34\displaystyle \int_{0}^q f(x) .dx = \frac{3}{4} but we can split it so that ∫01f(x).dx+∫1qf(x).dx=34\displaystyle \int_{0}^1 f(x) .dx + \int_1^q f(x) .dx = \frac{3}{4} and we know that ∫01f(x).dx=12\displaystyle \int_{0}^1 f(x) .dx = \frac{1}{2} so subtracting this from both sides yields the equality shown in the mark scheme.

d)
P(q<X<1.5)=P(X<1.5)−P(X<q)=[1−P(X>1.5)]−[1−P(X>q)]=P(X>q)−P(X>1.5)\begin{aligned}P(q<X<1.5) & = P(X < 1.5) - P(X<q) \\ & = [1-P(X>1.5)]-[1-P(X>q)] \\ & = P(X>q) - P(X>1.5) \end{aligned}

We want to preferably find P(X>1.5)P(X>1.5) because then our intervals of interest are strictly within the interval 1<x<21<x<2 so we only need to deal with a single function 14(5−2x)\frac{1}{4}(5-2x) (*). (Also since q is the upper quartile, we know that P(X>q) = 0.25 straight away without any work)

(*) If we didn't change our interval, then obviously we would need to deal with ∫01.5f(x).dx=∫01f(x).dx+∫11.5f(x).dx=12+∫11.5f(x).dx\displaystyle \int_{0}^{1.5} f(x) .dx = \int_0^1 f(x).dx + \int_1^{1.5} f(x) .dx = \dfrac{1}{2} + \int_{1}^{1.5} f(x) .dx and this isn't as bad since we know without thinking that the first one must 1/2, but in general this doesn't happen very often and you'd have to deal with two integrals instead of just one for a given piecewise pdf just as you have here.
(edited 6 years ago)

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