1. A bottling machine is calibrated to dispense 20 ounces of water to each bottl
ID: 3134676 • Letter: 1
Question
1.
A bottling machine is calibrated to dispense 20 ounces of water to each bottle. The actual
volume dispensed averages 20 ounces with standard deviation 0.2 ounces.
a.) If the distribution of individual bottle volumes is normal, what proportion of bottles receive
more than 20.39 ounces?
b.) What’s the probability that 2 of the 6 bottles in a six-pack have between 20.15 ounces and
20.25 ounces? (Again, assume that individual bottle volumes are normal in distribution.)
c.) If 36 bottles are randomly selected, what’s the probability that they average between 19.97
and 20.03 ounces in volume?
d.) If the standard deviation for individual bottle volumes was larger than 0.2 ounces, would your
answer in (c.) increase or decrease? Explain carefully – be sure to describe without any
supporting calculations...but be sure to refer to the impact of this change on the correct standard
deviation and its role.
e.) Would the answer to (c.) increase or decrease with a larger sample size? Again, explain.
For a. I got 2.56% of bottles by finding the z score with (20.39-20)/0.2 and plugging it in, and subtracting that percentage from 1. For b. I got 13.11% of bottles, since the probability of getting a bottle with the right amount should be 12.1% using a Z Table and the formula x(0.121^2)(0.879^4 where x is the 15 possible permutations with two correct bottles out of the 6 total). I don't know how to figure out c, so I cannot answer d and e. Any help would be much appreciated!
Explanation / Answer
MEAN = 20
STANDARD DEVIATION = 0.2
THE DISTRIBUTION IS GIVEN TOBE NORMAL
THEREFORE WE WILL APPLY THE Z TEST FORMULA
Z SCORE = (X-MEAN)/STANDARD DEVIATION
A)P(X>20.39) =
For x = 20.39, z = (20.39 - 20) /0.2 = 1.95
Hence P(x > 20.39) = P(z > 1.95) = [total area] - [area to the left of 1.95]
1 - [area to the left of 1.95]
now from the z table we will take the value of z score = 1.95
= 1 - 0.9744 = 0.0256
THEREFOR PROPORTION WILL BE 2.56%
B)1ST WE NEED TO FIND THE PROBABILITY BETWEEN P(20.15<X<20.25)
NOW WE GET
) For x = 20.15 , z = (20.15 - 20) / 0.2 = 0.75 and for x = 20.25, z = (20.25 - 20) / 0.2 = 0.8
Hence P(20.15 < x < 20.25) = P(0.75 < z < 0.8) = [area to the left of z = 0.8] - [area to the left of 0.75]
= 0.7881 - 0.7734 = 0.0147
SO TO GET 2 OUT OF 6
= 6C2*(0.0147)^2*(0.9853)^4 = 0.0030
C)
1ST WE NEED TO FIND THE PROBABILITY BETWEEN P(19.97<X<20.03)
NOW WE GET
) For x = 19.97 , z = (19.97 - 20) / 0.2 = -0.15 and for x = 20.03, z = (20.03 - 20) / 0.2 = 0.15
Hence P(19.97 < x < 20.03) = P(-0.15 < z < 0.15) = [area to the left of z = 0.15] - [area to the left of -0.15]
= 0.5596 - 0.4404 = 0.1192
D) IF THE VALUE OF THE STANDARD DEVIATION WILL BE MORE THEN 0.2 THEN THE Z VALUE FOR THE BOTH THE VALUE 19.97 AND THE 20.03 WOULD HVE DECRESED WHICH ULTIMATELY HAVE REDUCED OR DECREASED THE ANSWER.
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