Question

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.

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Answer:

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 $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 1.96*\frac{0.15}{\sqrt{36}} = 0.049$

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.