An automatic filling machine is used to fill two-litre bottles of cola. The machine"s output is know to be approximately normal with a mean of 2.0 litres and a standard deviation that .01 litres. Output is monitored using means of samples of five observations. a. determine the upper and lower limits that will include roughly 95.5% of the sample means. b. if the means for 6 samples are: 2.005, 2.001 1.998, 2.002 1.995, and 1.999, is the process in control?
Who are the experts?Our certified Educators are real professors, teachers, and scholars who use their academic expertise to tackle your toughest questions. Educators go through a rigorous application process, and every answer they submit is reviewed by our in-house editorial team.

You are watching: An automatic filling machine is used to fill 1-liter bottles of cola


*

The output is approximately normal with `mu=2,sigma=.01` and tested using 5 observations.

(a) Determine the upper and lower bounds that will include approximately 95.5% of the sample means.

Since we know the population standard deviation we can use a z-interval. The bounds are found by:

`2 +- z_(alpha/2)(sigma/sqrt(n))`

Here `alpha=.045...


Start your 48-hour free trial to unlock this answer and thousands more. Enjoy tacoemojishirt.com ad-free and cancel anytime.


The output is approximately normal with `mu=2,sigma=.01` and tested using 5 observations.

(a) Determine the upper and lower bounds that will include approximately 95.5% of the sample means.

See more: Super 3D Space Jam World 64 Roblox Song Id, Pin On Bgm I Liked

Since we know the population standard deviation we can use a z-interval. The bounds are found by:

`2 +- z_(alpha/2)(sigma/sqrt(n))`

From a standard normal table or technology we find `z_(alpha/2)~~2`

** If using a table you will uncover .0225 corresponds to a z-value of -2. Since the table is symmetric we can use the positive value. Otherwise look up 1-.0225=.9775 to find z=2 **

`sigma=.01,n=5` so substituting the values we get:

`2-2(.01/sqrt(5))~~1.9911` and `2+2(.01/sqrt(5))~~2.0089`

------------------------------------------------

So the bounds are 1.9911 and 2.0089

------------------------------------------------

** My calculator gives the interval as 1.991,2.009 **

(b) The process is in control as long as the data points lie within the bounds. All of the given data points are within the upper and lower bounds, so the process is in control.