probability involving z-score

• Jul 26th 2010, 06:45 PM
Jskid
[RESOLVED]probability involving z-score
A bottling company uses a machine to fill plastic bottles with pop. Due to variations in the filling process, the pop amounts in all bottles are normally distributed with a mean of 300 ml and a standard deviation of 3 ml. One bottle is randomly selected, what is the probability that bottle contains less than 299.5 ml?

$\displaystyle z=\frac{299.5-300}{\frac{3}{\sqrt{n}}}$ I'm not sure what value n is, or if it's needed.

I think I'm suppose to take n=1.
For this question how I'm calculating the probability do I write $\displaystyle P(\bar{x}<299.5)$ or $\displaystyle P(x<299.5)$?
• Jul 26th 2010, 08:20 PM
mr fantastic
Quote:

Originally Posted by Jskid
A bottling company uses a machine to fill plastic bottles with pop. Due to variations in the filling process, the pop amounts in all bottles are normally distributed with a mean of 300 ml and a standard deviation of 3 ml. One bottle is randomly selected, what is the probability that bottle contains less than 299.5 ml?

$\displaystyle z=\frac{299.5-300}{\frac{3}{\sqrt{n}}}$ I'm not sure what value n is, or if it's needed.

Where has n come from? $\displaystyle \displaystyle z = \frac{x - \mu}{\sigma}$ and you have been told the values of mean, sd and x.
• Jul 27th 2010, 05:26 PM
Jskid
$\displaystyle \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$ where n is sample size.
What I don't understand is a remeber doing similar questions but just used the S.D. $\displaystyle \sigma$
• Jul 27th 2010, 05:42 PM
pickslides
Quote:

Originally Posted by Jskid
$\displaystyle \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$ where n is sample size.
What I don't understand is a remeber doing similar questions but just used the S.D. $\displaystyle \sigma$

$\displaystyle P(\bar{X}<299.5) \implies P\left(Z<\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

$\displaystyle P(X<299.5) \implies P(Z<\frac{X-\mu}{\sigma})$

Can you see the difference?
• Jul 27th 2010, 07:22 PM
Jskid
Quote:

Originally Posted by pickslides
$\displaystyle P(\bar{X}<299.5) \implies P\left(Z<\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

$\displaystyle P(X<299.5) \implies P(Z<\frac{X-\mu}{\sigma})$

Can you see the difference?

I think I understand. The top equation is for the probability of a sample mean and the lower one is for the probability of getting a sample of a certain value? Does that mean the two formula are the same because $\displaystyle \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{1}}}=\frac{X-\mu}{\frac{\sigma}{1}}$
• Jul 27th 2010, 09:18 PM
pickslides
Quote:

Originally Posted by Jskid
I think I understand. The top equation is for the probability of a sample mean and the lower one is for the probability of getting a sample of a certain value? Does that mean the two formula are the same because $\displaystyle \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{1}}}=\frac{X-\mu}{\frac{\sigma}{1}}$

I have never thought of it like that. But I am inclinded to say this is not true as you can be given a sample size $\displaystyle n>1$ for $\displaystyle X$ and the standard tranformation would apply given it was normal.