Turn on thread page Beta
    • Thread Starter
    Offline

    13
    ReputationRep:
    Please could someone explain questions 6b and c

    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

    Paper: http://filestore.aqa.org.uk/sample-p...W-QP-JUN11.PDF

    Mark scheme: http://filestore.aqa.org.uk/sample-p...W-MS-JUN11.PDF
    • Community Assistant
    Online

    20
    ReputationRep:
    Community Assistant
    (Original post by jazz_xox_)
    Please could someone explain questions 6b and c

    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

    Paper: http://filestore.aqa.org.uk/sample-p...W-QP-JUN11.PDF

    Mark scheme: http://filestore.aqa.org.uk/sample-p...W-MS-JUN11.PDF
    b) This is simply from the fact that the median, m, is a point along the distribution of X, F_X(x), so that F_X(m) = \frac{1}{2} (i.e. the middle value of the distribution) - so then just plug in x=1 to obtain F(1)=\frac{1}{2} and this verifies that it's the median.

    c) This is because we have \displaystyle \int_{0}^q f(x) .dx = \frac{3}{4} but we can split it so that \displaystyle \int_{0}^1 f(x) .dx + \int_1^q f(x) .dx = \frac{3}{4} and we know that \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)
    \begin{aligned}P(q&lt;X&lt;1.5) &  = P(X &lt; 1.5) - P(X&lt;q) \\ & = [1-P(X&gt;1.5)]-[1-P(X&gt;q)] \\ & = P(X&gt;q) - P(X&gt;1.5) \end{aligned}

    We want to preferably find P(X&gt;1.5) because then our intervals of interest are strictly within the interval 1&lt;x&lt;2 so we only need to deal with a single function \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 \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.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 13, 2018

University open days

  1. University of Cambridge
    Christ's College Undergraduate
    Wed, 26 Sep '18
  2. Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 28 Sep '18
  3. Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 29 Sep '18
Poll
Which accompaniment is best?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.