Normal distribution question

Homework Statement

3 (a) A sample of 50 washed baking potatoes was selected at random from a large batch.
The weights of the 50 potatoes were found to have a mean of 234 grams and a standard
deviation of 25.1 grams.
Construct a 95% confidence interval for the mean weight of potatoes in the batch.
(4 marks)

Homework Equations

$\bar{x} = \mu$

$s = \frac{\sigma}{\sqrt{n}}$

$z = \frac{x-\mu}{s}$

The Attempt at a Solution

The confidence interval 95% means p = 0.975

$z = \frac{x-\mu}{s}$

$\pm 1.96 = \frac{x-234}{\frac{25.1}{\sqrt{50}}}$

$\Rightarrow 227-241$

Why is the standard deviation

$s = \frac{25.1}{\sqrt{50}}$

and not just 25.1. The question states that the s.d. is 25.1!

Related Precalculus Mathematics Homework Help News on Phys.org
$\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$ is approximately N(0,1) for large n (assuming that the original distribtuion is not skewed)

Mark44
Mentor
Why is the standard deviation

$s = \frac{25.1}{\sqrt{50}}$

and not just 25.1. The question states that the s.d. is 25.1!
There are two things going on here. On the one hand there are the population standard deviation ($\sigma$ and sample standard deviation s. On the other is the standard deviation of the mean, which is defined as:
$$\sigma_{mean} = \frac{\sigma}{\sqrt{n}}$$

The wikipedia article here--http://en.wikipedia.org/wiki/Standard_deviation--talks [Broken] about the st. dev. of the mean in the section titled Relationship between standard deviation and mean.

Last edited by a moderator:
There are two things going on here. On the one hand there are the population standard deviation ($\sigma$ and sample standard deviation s. On the other is the standard deviation of the mean, which is defined as:
$$\sigma_{mean} = \frac{\sigma}{\sqrt{n}}$$

The wikipedia article here--http://en.wikipedia.org/wiki/Standard_deviation--talks [Broken] about the st. dev. of the mean in the section titled Relationship between standard deviation and mean.
That's what's confusing, the standard deviation of the mean is the same as the standard deviation of the sample divided by root n?

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That's what's confusing, the standard deviation of the mean is the same as the standard deviation of the sample divided by root n?
It's approximately the same.

s =SIGMA/SQRT(N)

STANDS FOR STANDARD ERROR NOT FOR STANDARD DEVAITION!!!!!!!!

s =SIGMA/SQRT(N)

STANDS FOR STANDARD ERROR NOT FOR STANDARD DEVAITION!!!!!!!!
The standard deviation of the sample mean is sometimes called the standard error.