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quick S3 normal distribution question

If I have a sample mean normally distributed by
Unparseable latex formula:

\[\bar{X}\sim N(\mu ,\frac{\sigma ^{2}}{n})

, how would I find
Unparseable latex formula:

P(\left | \bar{X} - \mu \right | < 15)\]

? Or better yet, how would I standardise the z-value?
Original post by aymanzayedmannan
If I have a sample mean normally distributed by
Unparseable latex formula:

\[\bar{X}\sim N(\mu ,\frac{\sigma ^{2}}{n})

, how would I find
Unparseable latex formula:

P(\left | \bar{X} - \mu \right | < 15)\]

? Or better yet, how would I standardise the z-value?


If mod (xbar - mu) is less than 15 then either one of two inequalities are satisfied.. what are they?
Original post by aymanzayedmannan
If I have a sample mean normally distributed by
Unparseable latex formula:

\[\bar{X}\sim N(\mu ,\frac{\sigma ^{2}}{n})

, how would I find
Unparseable latex formula:

P(\left | \bar{X} - \mu \right | < 15)\]

? Or better yet, how would I standardise the z-value?


If you have

XˉN(μ,σ2n) \displaystyle \bar{X} \sim N\left(\mu ,\frac{\sigma ^{2}}{n}\right)

then,

XˉμN(0,σ2n) \displaystyle \bar{X} - \mu \sim N\left(0 ,\frac{\sigma ^{2}}{n}\right)

and then

n(Xˉμ)σN(0,1) \displaystyle \frac{\sqrt{n} \left(\bar{X} - \mu \right)}{\sigma} \sim N(0 ,1)

Clearer now?
Reply 3
Original post by SeanFM
If mod (xbar - mu) is less than 15 then either one of two inequalities are satisfied.. what are they?


Unparseable latex formula:

\[P(-15 < \bar{X}- \mu < 15)\]

, right?


Original post by Gregorius
If you have

XˉN(μ,σ2n) \displaystyle \bar{X} \sim N\left(\mu ,\frac{\sigma ^{2}}{n}\right)

then,

XˉμN(0,σ2n) \displaystyle \bar{X} - \mu \sim N\left(0 ,\frac{\sigma ^{2}}{n}\right)

and then

n(Xˉμ)σN(0,1) \displaystyle \frac{\sqrt{n} \left(\bar{X} - \mu \right)}{\sigma} \sim N(0 ,1)

Clearer now?


I can understand how you got there by way of expectation algebra. would you just substitute the given values from here and then use a standardised normal distribution?
(edited 8 years ago)
Original post by aymanzayedmannan
Unparseable latex formula:

\[P(-15 < \bar{X}- \mu < 15)\]

, right?




I can understand how you got there by way of expectation algebrawould you just substitute the given values from here and then use a standardised normal distribution?


Mhm, but the method that Gregorius has suggested is much neater. :colondollar:
(edited 8 years ago)
Original post by SeanFM
Mhm, but the method Gregorius has suggest is much neater. :colondollar:


Flattery will bring you much rep. :smile:
Reply 6
Original post by SeanFM
Mhm, but the method that Gregorius has suggested is much neater. :colondollar:


Original post by Gregorius
Flattery will bring you much rep. :smile:


It certainly is elegant; cheers, guys!
Original post by aymanzayedmannan

I can understand how you got there by way of expectation algebrawould you just substitute the given values from here and then use a standardised normal distribution?


If you have

15<Xˉμ<15\displaystyle -15 < \bar{X}- \mu < 15,

then you can say something about the bounds satisfied by

n(Xˉμ)σ \displaystyle \frac{\sqrt{n} \left(\bar{X} - \mu \right)}{\sigma}

As you know that this quantity is distributed as a standard normal variable, you can then work out the requisite probability.
Reply 8
Original post by Gregorius
If you have

15<Xˉμ<15\displaystyle -15 < \bar{X}- \mu < 15,

then you can say something about the bounds satisfied by

n(Xˉμ)σ \displaystyle \frac{\sqrt{n} \left(\bar{X} - \mu \right)}{\sigma}

As you know that this quantity is distributed as a standard normal variable, you can then work out the requisite probability.


Thank you for the help, Gregorius :smile: This has cleared a lot of concepts while self-teaching.

sorry for bringing this up again - I looked back at this while I was working through a few problems

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