Question

A manufacturer of cheese filled ravioli supplies a pizza restaurant chain. Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed. To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 ravioli and measure the weight of cheese filling. They find a sample mean weight of 15 grams with a standard deviation of 1.5 grams. What is the standard deviation of sample mean

Answer 1

Answer:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard deviation for the sample mean would be given by:

$\sigma_{\bar X}= \frac{1.5}{\sqrt{25}}= 0.3$

Step-by-step explanation:

For this case we know that the amount of cheese inserted into the ravioli is normally distributed. And we have the following info given;

$\bar X =15$ the sample mean

$s= 1.5$ the sample deviation

$n=25$ the sample size

And for this case we know that the sample size is large enough in order to apply the central limit theorem and the distribution for the sample mean would be given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard deviation for the sample mean would be given by:

$\sigma_{\bar X}= \frac{1.5}{\sqrt{25}}= 0.3$