A soft-drink machine is regulated so that the amount of drink dispensed is approximately normally distributed with standard deviation equal to 0.15 deciliter. Find a 95 % confidence level for the mean of all drinks dispensed by this machine if a random sample of 36 drinks has an average content of 2.25 deciliters.

The 95 % confidence level for the mean of all drinks dispensed by this machine is between 2.2010 deciliters and 2.2990 deciliters.

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 M as such

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

The lower end of the interval is the mean subtracted by M. So it is 2.25 - 0.049 = 2.2010 deciliters

The upper end of the interval is the mean added to M. So it is 2.25 + 0.049 = 2.2990 deciliters

The 95 % confidence level for the mean of all drinks dispensed by this machine is between 2.2010 deciliters and 2.2990 deciliters.

step-by-step explanation:

the correct option is b

step-by-step explanation: