The Student Room Group

quick s3 question

Real quick question for anyone interested :colondollar:

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 :s-smilie:
Original post by Katiee224
Real quick question for anyone interested :colondollar:

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 :s-smilie:


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.
Reply 2
Original post by Katiee224
Real quick question for anyone interested :colondollar:

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 :s-smilie:


We know that TˉN(μ,2.52n)\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:

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

Var(Tˉμ)=2.52n\displaystyle \mathrm{Var} \left ( \bar{T} - \mu \right ) = \frac{2.5^{2}}{n} [\because the mean, μ\mu, has no variance]

(Tˉμ)N(0,2.52n)\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 XX. :h:

@Zacken, how do you do these, out of interest?
(edited 8 years ago)
Reply 3
Original post by aymanzayedmannan


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


TˉN(μ,2.52n)\displaystyle \bar{T} \sim N\left(\mu, \frac{2.5^2}{n}\right) so we want P((Tˉμ<12)=0.95\mathbb{P}\left((|\bar{T} - \mu| < \frac{1}{2}\right) = 0.95

Hence, since P(Tˉμ<12)=2P(Tˉμ<12)1\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: P(Tˉ<12+μ)=0.975\displaystyle \mathbb{P}\left(\bar{T} < \frac{1}{2} + \mu\right) = 0.975

Standardising gives us P(Z<0.5+μμ2.5n)=0.975P((Z>0.5n2.5)=0..025\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 0.5n2.5=1.96\frac{0.5\sqrt{n}}{2.5} = 1.96 from the percentage points table.
Reply 4
Original post by aymanzayedmannan
We know that TˉN(μ,2.52n)\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:

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

Var(Tˉμ)=2.52n\displaystyle \mathrm{Var} \left ( \bar{T} - \mu \right ) = \frac{2.5^{2}}{n} [\because the mean, μ\mu, has no variance]

(Tˉμ)N(0,2.52n)\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 XX. :h:

@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?:colondollar: does that just equal 0?
(edited 8 years ago)
Reply 5
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?:colondollar: does that just equal 0?


E(Tˉμ)=E(Tˉ)μ\displaystyle E\left(\bar{T} - \mu\right) = E(\bar{T}) - \mu using E(X+b)=E(x)+bE(X + b) = E(x) + b, but since E(Tˉ)=μE(\bar{T}) = \mu then we have E(Tˉμ)=μμ=0E(\bar{T} - \mu) = \mu - \mu = 0.

We also have Var(X+b)=Var(X)\mathrm{Var}(X+b) = \mathrm{Var}(X) so Var(Tˉμ)=Var(Tˉ)=2.52n\mathrm{Var}(\bar{T} - \mu) = \mathrm{Var}(\bar{T}) = \frac{2.5^2}{n}.
Reply 6
Original post by Zacken
E(Tˉμ)=E(Tˉ)μ\displaystyle E\left(\bar{T} - \mu\right) = E(\bar{T}) - \mu using E(X+b)=E(x)+bE(X + b) = E(x) + b, but since E(Tˉ)=μE(\bar{T}) = \mu then we have E(Tˉμ)=μμ=0E(\bar{T} - \mu) = \mu - \mu = 0.

We also have Var(X+b)=Var(X)\mathrm{Var}(X+b) = \mathrm{Var}(X) so Var(Tˉμ)=Var(Tˉ)=2.52n\mathrm{Var}(\bar{T} - \mu) = \mathrm{Var}(\bar{T}) = \frac{2.5^2}{n}.


Okay so you are saying we treat μ \mu as a constant?
Reply 7
Original post by Katiee224
Okay so you are saying we treat μ \mu as a constant?


We treat it as a constant because it is a constant.
Reply 8
Original post by Katiee224
Okay so you are saying we treat μ \mu as a constant?


The mean has no variance, so it is a constant.
Reply 9
okay i feel a bit stupid now haha, thanks for clearing that one up foe me guys :biggrin::biggrin:
Reply 10
Original post by Katiee224
okay i feel a bit stupid now haha, thanks for clearing that one up foe me guys :biggrin::biggrin:


Nothing to feel stupid about, it's good that you're asking these things. :biggrin:

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