Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    11
    ReputationRep:
    Real quick question for anyone interested

    why isn't the fourth line following the usual normal distribution layout of the random variable minus the mean all over the standard deviation? so it should be 0.5 - the mean?

    so confused with what's going on in this question
    Attached Images
     
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by Katiee224)
    Real quick question for anyone interested

    why isn't the fourth line following the usual normal distribution layout of the random variable minus the mean all over the standard deviation? so it should be 0.5 - the mean?

    so confused with what's going on in this question
    It's not quite what they're doing in this question.

    Between lines 3 and 4, they've divided both sides by the standard deviation of the variable tbar. It transforms the left hand side inside the brackets into Z and the right hand side into a function of n and then you use that to find n.
    Offline

    3
    ReputationRep:
    (Original post by Katiee224)
    Real quick question for anyone interested

    why isn't the fourth line following the usual normal distribution layout of the random variable minus the mean all over the standard deviation? so it should be 0.5 - the mean?

    so confused with what's going on in this question
    We know that \displaystyle \bar{T}\simeq \sim \mathrm{N}\left ( \mu, \frac{2.5^{2}}{n} \right ).

    The way I like to do it is by finding a new distribution like this:

    \displaystyle \mathrm{E}\left ( \bar{T} - \mu \right ) = 0

    \displaystyle \mathrm{Var} \left ( \bar{T} - \mu \right ) = \frac{2.5^{2}}{n} [\because the mean, \mu, has no variance]

    \displaystyle \Rightarrow \left ( \bar{T} - \mu\right) \sim \mathrm{N}\left ( 0, \frac{2.5^{2}}{n} \right )

    and then we base our calculations off of the new distribution we found. Might even help to call it X.

    Zacken, how do you do these, out of interest?
    Offline

    22
    ReputationRep:
    (Original post by aymanzayedmannan)

    Zacken, how do you do these, out of interest?
    \displaystyle \bar{T} \sim N\left(\mu, \frac{2.5^2}{n}\right) so we want \mathbb{P}\left((|\bar{T} - \mu| < \frac{1}{2}\right) = 0.95

    Hence, since \displaystyle \mathbb{P}\left(|\bar{T} - \mu| < \frac{1}{2}\right) = 2\mathbb{P}\left(\bar{T} - \mu < \frac{1}{2}\right) - 1

    Which yields: \displaystyle \mathbb{P}\left(\bar{T} < \frac{1}{2} + \mu\right) = 0.975

    Standardising gives us \displaystyle \mathbb{P}\left(Z < \frac{0.5 + \mu - \mu}{\frac{2.5}{\sqrt{n}}}\right  ) = 0.975 \Rightarrow \mathbb{P}\left((Z > \frac{0.5\sqrt{n}}{2.5}\right) = 0..025

    So we know that \frac{0.5\sqrt{n}}{2.5} = 1.96 from the percentage points table.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by aymanzayedmannan)
    We know that \displaystyle \bar{T}\simeq \sim \mathrm{N}\left ( \mu, \frac{2.5^{2}}{n} \right ).

    The way I like to do it is by finding a new distribution like this:

    \displaystyle \mathrm{E}\left ( \bar{T} - \mu \right ) = 0

    \displaystyle \mathrm{Var} \left ( \bar{T} - \mu \right ) = \frac{2.5^{2}}{n} [\because the mean, \mu, has no variance]

    \displaystyle \Rightarrow \left ( \bar{T} - \mu\right) \sim \mathrm{N}\left ( 0, \frac{2.5^{2}}{n} \right )

    and then we base our calculations off of the new distribution we found. Might even help to call it X.

    Zacken, how do you do these, out of interest?
    I'm still a bit lost, how does this follow the rules of E(X-Y) = E(X) - E(Y) and Var(X-Y) = Var(X) + Var(Y)?

    like how can you have E(x) with x being another mean? does that just equal 0?
    Offline

    22
    ReputationRep:
    (Original post by Katiee224)
    I'm still a bit lost, how does this follow the rules of E(X-Y) = E(X) - E(Y) and Var(X-Y) = Var(X) + Var(Y)?

    like how can you have E(x) with x being another mean? does that just equal 0?
    \displaystyle E\left(\bar{T} - \mu\right) = E(\bar{T}) - \mu using E(X + b) = E(x) + b, but since E(\bar{T}) = \mu then we have E(\bar{T} - \mu) = \mu - \mu = 0.

    We also have \mathrm{Var}(X+b) = \mathrm{Var}(X) so \mathrm{Var}(\bar{T} - \mu) = \mathrm{Var}(\bar{T}) = \frac{2.5^2}{n}.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by Zacken)
    \displaystyle E\left(\bar{T} - \mu\right) = E(\bar{T}) - \mu using E(X + b) = E(x) + b, but since E(\bar{T}) = \mu then we have E(\bar{T} - \mu) = \mu - \mu = 0.

    We also have \mathrm{Var}(X+b) = \mathrm{Var}(X) so \mathrm{Var}(\bar{T} - \mu) = \mathrm{Var}(\bar{T}) = \frac{2.5^2}{n}.
    Okay so you are saying we treat  \mu as a constant?
    Offline

    22
    ReputationRep:
    (Original post by Katiee224)
    Okay so you are saying we treat  \mu as a constant?
    We treat it as a constant because it is a constant.
    Offline

    3
    ReputationRep:
    (Original post by Katiee224)
    Okay so you are saying we treat  \mu as a constant?
    The mean has no variance, so it is a constant.
    • Thread Starter
    Offline

    11
    ReputationRep:
    okay i feel a bit stupid now haha, thanks for clearing that one up foe me guys
    Offline

    22
    ReputationRep:
    (Original post by Katiee224)
    okay i feel a bit stupid now haha, thanks for clearing that one up foe me guys
    Nothing to feel stupid about, it's good that you're asking these things.
 
 
 
Poll
Who is your favourite TV detective?
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.