# I think this is about the Central Limit Theorem

#### whitejac

1. Homework Statement
An engineer is measuring a quantity q. It is assumed that there is a random error in each measurement, so the engineer will take n measurements and reports the average of the measurements as the estimated value of q. Specifically, if Yi is the value that is obtained in the i'th measurement, we assume that

Yi=q+Xi,
where Xi is the error in the ith measurement. We assume that Xi's are i.i.d. with EXi=0 and Var(Xi)=4 units. The engineer reports the average of measurements

Mn = (Y1 + .... Yn) / n

How
many measurements does the engineer need to make until he is 95% sure that the final error is less than 0.1 units? In other words, what should the value of n be such that
P(q−0.1≤Mn≤q+0.1)≥0.95?

2. Homework Equations
Central Limit Theorem states:
Zn = (Mx - μ) / (σ / √n)

3. The Attempt at a Solution
So this is the formula I chose to use. It seems like a simple variable swap, but my problem is pulling n out.

P(y1 ≤ Y ≤ y2)
= P( (y1 - nμ) / (σ√n) ≤ ((X1 +... Xn) - nμ) / (σ√n)) ≤ (y2 - nμ) / (σ√n)

Then this would give ((q + 0.1) / (2√n)) - ((q - 0.1) / (2√n)) = Φ-1 (0.95) = 1.64

combining this would give:(q/2√n) + (0.1 / 2√n) - (q / 2√n)- (0.1 / 2√n) = 0.2 / 2√n = 1/√n = 1.64 ⇒ n = (1/1.64)2 = 0.37

Now... this is not an integer, and makes absolutely no sense. I am semi-confident in my process, but I think I may have made too many assumptions about the central limit theorem.

Related Calculus and Beyond Homework Help News on Phys.org

#### BvU

Homework Helper
Why do you change from $\ \sigma\over\sqrt n\$ to $\ \sigma\sqrt n\$ ?

It seems your n is used for two different purposes as well

...

#### whitejac

Why do you change from $\ \sigma\over\sqrt n\$ to $\ \sigma\sqrt n\$ ?

It seems your n is used for two different purposes as well

...
When you multiply by n/n you get σ√n

#### BvU

Homework Helper
I would call that multiplying by n, not multiplying by n/n

P(y1 ≤ Y ≤ y2)
= P( (y1 - nμ) / (σ√n) ≤ ((X1 +... Xn) - nμ) / (σ√n)) ≤ (y2 - nμ) / (σ√n)

Then this would give ((q + 0.1) / (2√n)) - ((q - 0.1) / (2√n)) = Φ-1 (0.95) = 1.64
Check it anyway !

#### whitejac

Central Limit Theorem states:
Zn = (Mx - μ) / (σ / √n)
It has to be n/n because Mx is (X1 +...+Xn)/n.

In my extrapolation of the formula you quoted, the y1 = q - 0.1, nμ = 0 and σ = √4
Which portion did I misinterpret/use twice?

#### BvU

Homework Helper
The "Then" part, where you divide by n

Another comment: Your 1.64 looks one-sided to me. And your notation asks more from the reader than is reasonable; teachers shouldn't allow that ! (there is no using n twice; I got confused -- sorry)

I end up with an awful lot of measurements required, which is rather to be expected if you start out with such a large variance !

By the way, even "Var(Xi)=4 units" is confusing. Shouldn't it be Var(Xi)=4 units2 ?

Central Limit Theorem states: Zn = (Mx - μ) / (σ / √n) is normally distributed with standard deviation 1

Matter of completeness: for some readers Zn is something else: an impedance or whatever. Last edited:

#### whitejac

By the way, even "Var(Xi)=4 units" is confusing. Shouldn't it be Var(Xi)=4 units2 ?
Sorry, it ought to be. My book isn't the best despite my professor's claim. I guess they're assuming "units" is general enough.
The "Then" part, where you divide by n
Maybe I'm not looking at it the same way you are, I don't divide by n. I multiply by n to get y1 and y2 = q-0.1, q+0.1, respectively. This is consistent with examples in my book, I think. Are you saying I mistook these values as y1,y2 when they were in fact y1 - nμ, y2 + nμ?

I apologize for my not showing as many steps, or in the best way. It's much neater on my paper. I'll upload it when i get home if it remains unclear.

#### BvU

Homework Helper
No need to apologize, I'm trying to help.
P(y1 ≤ Y ≤ y2)
= P( (y1 - nμ) / (σ√n) ≤ ((X1 +... Xn) - nμ) / (σ√n)) ≤ (y2 - nμ) / (σ√n)

Then this would give ((q + 0.1) / (2√n)) - ((q - 0.1) / (2√n)) = Φ-1 (0.95) = 1.64
In the top line you have y1 for nq, in the bottom line you have q only. You only divide the numerator by n.

The 1.64 is for 5% in one tail, you want 2.5% in each tail.

--

 The more I read it, the more nonsensical it becomes:

P(y1 ≤ Y ≤ y2) = P( (y1 - nμ) / (σ√n) ≤ ((X1 +... Xn) - nμ) / (σ√n)) ≤ (y2 - nμ) / (σ√n)
What does it say here ?

Last edited:

#### Ray Vickson

Homework Helper
Dearly Missed
Sorry, it ought to be. My book isn't the best despite my professor's claim. I guess they're assuming "units" is general enough.

Maybe I'm not looking at it the same way you are, I don't divide by n. I multiply by n to get y1 and y2 = q-0.1, q+0.1, respectively. This is consistent with examples in my book, I think. Are you saying I mistook these values as y1,y2 when they were in fact y1 - nμ, y2 + nμ?

I apologize for my not showing as many steps, or in the best way. It's much neater on my paper. I'll upload it when i get home if it remains unclear.
I suggest you go back to square one. You need to figure out how to make a 95% probability interval (centered at the mean) come out with a length of 0.2 (i.e., from -0.1 to + 0.1). In terms of the unit normal distribution, how many standard deviations wide would that be? Then carefully figure out what that means in terms of n. When I do it I get a value of n between 1000 and 2000, but I will not state the precise value.