Central Limit Theorem

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June 17th 2009, 07:56 PM
shinn

Central Limit Theorem

Hi,

I have a problem regarding CLT:

"The lengths of mass produced nails are normally distributed with a mean of $3cm$ and variance of $0.01 cm^2$ . 16 nails are randomly chosen and laid end to end.

What is the probability that its length exceeds 48.5cm?"

This is my working out, I'm not sure what i am doing wrong:

Let $T_{16} = X_1 + X_2 + .... + X_{16}$ , where $X_i$ are identically and independently distributed variables.

So, each of $X_i$ ~ $N(3,0.01^2)$
Thus, $T_n$ ~ $N(48, 0.04^2)$

We need to find $P(T_{16} > 48.5)$ . So
$P(T_{16} > 48.5) = 1 - P( T_{16} < 48.5) = 1 - P( Z < (48.5 - 48) / (0.04) ) = 1 - P( Z < 12.5)$

Then, I'm stuck here. Any help will be greatly appreciated thanks(Itwasntme)
June 17th 2009, 08:02 PM
mr fantastic

Quote:

Originally Posted by shinn

Hi,

I have a problem regarding CLT:

"The lengths of mass produced nails are normally distributed with a mean of $3cm$ and variance of $0.01 cm^2$ . 16 nails are randomly chosen and laid end to end.

What is the probability that its length exceeds 48.5cm?"

This is my working out, I'm not sure what i am doing wrong:

Let $T_{16} = X_1 + X_2 + .... + X_{16}$ , where $X_i$ are identically and independently distributed variables.

So, each of $X_i$ ~ $N(3,0.01^2)$
Thus, $T_n$ ~ $N(48, 0.04^2)$

We need to find $P(T_{16} > 48.5)$ . So
$P(T_{16} > 48.5) = 1 - P( T_{16} < 48.5) = 1 - P( Z < (48.5 - 48) / (0.04) ) = 1 - P( Z < 12.5)$

Then, I'm stuck here. Any help will be greatly appreciated thanks(Itwasntme)

The sum follows a normal distribution.

The mean of the sum is 48. The variance of the sum is (16)(0.01) = 0.16 therefore the sd of the sum is 0.4.
June 17th 2009, 08:33 PM
matheagle

There are a few errors here.
1. There isn't a CLT here, the sample is quite small.
So no limit theorem whatsoever.
BUT you are told that the 16 X's are normal, hence their sum is normal.
2. The variance is .01 not $(.01)^2$ .
3. The st deviation of $T_n$ is also wrong, it is .4.

$P(T_{16} > 48.5) = 1 - P( T_{16} < 48.5) = 1 - P( Z < (48.5 - 48) / .4 ) = 1 - P( Z < 1.25)$
is what you want to look up.
June 17th 2009, 09:27 PM
shinn

thanks for the above replies.