Understanding 95% Confidence Interval & Total Width <2.0

[inv]

[Solved]Is a stats question,just need some clarity on it-"total width of less than 2.

Homework Statement

Hi.The result of a memory test is known to be normally distributed with mean mew and standard deviation 1.0. It's required to have a 95% confidence interval for mew with a total width of less than 2.0.Find the least possible number of tests needed to achieve this.*The "total width of less than 2.0" mean what?

Homework Equations

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The Attempt at a Solution

P(Z<2.0)=0.95
(X-mew)/(1.9/n^.5)<2.0,where X=the mean
mew<2.0
I'm stuck here,I'm sure something is wrong with it.So pls explain the thing with the * above?

 

[HallsofIvy]

Homework Statement

Hi.The result of a memory test is known to be normally distributed with mean mew and standard deviation 1.0. It's required to have a 95% confidence interval for mew with a total width of less than 2.0.Find the least possible number of tests needed to achieve this.*The "total width of less than 2.0" mean what?

Any confidence interval for the true mean is of the form [a,b]. The "total width" is b-a. In particular, if you take the interval to be symmetric about mew (since the normal distribution is symmetric about its mean) then you want [mew- a,mew+ a] with "total width"= 2a< 2 or a< 1.

Homework Equations

?

The Attempt at a Solution

P(Z<2.0)=0.95
(X-mew)/(1.9/n^.5)<2.0,where X=the mean
mew<2.0
I'm stuck here,I'm sure something is wrong with it.So pls explain the thing with the * above?

No, You want P(mew-1< X< mew+1)= 0.95. (X-mew)/(1.9/n^.5) with X= mew+ 1 gives z= 1/(1.9/n^.5)= n^.5/1.9. Look up the z score that gives .95/2 = .475 (for a table that uses only positive values) and then solve for n.

 

[inv]

Thx.Problem solved~The End~