# Normal distribution question

1. May 15, 2009

### Gregg

1. The problem statement, all variables and given/known data

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)

2. Relevant equations

$\bar{x} = \mu$

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

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

3. 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!

2. May 15, 2009

### Random Variable

$\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$ is approximately N(0,1) for large n (assuming that the original distribtuion is not skewed)

3. May 15, 2009

### Staff: Mentor

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: May 4, 2017
4. May 15, 2009

### Gregg

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?

Last edited by a moderator: May 4, 2017
5. May 15, 2009

### Random Variable

It's approximately the same.

6. May 16, 2009

### rclakmal

s =SIGMA/SQRT(N)

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

7. May 16, 2009

### Gregg

The standard deviation of the sample mean is sometimes called the standard error.