You measure 40 watermelons' weights, and find they have a mean weight of 66 ounces. Assume the population standard deviation is 13.3 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight.

The maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight is of 3.46 ounces.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

In this problem:

So

The maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight is of 3.46 ounces.

v = 1/3 a^2h

step-by-step explanation:

if the height is 57.7cm and the height of the slant edge is 18.2cm, then the volume is 1/3 a^2h.