Q3 A bottling machine can be regulated so that it discharges an average of pu ou
ID: 3023394 • Letter: Q
Question
Q3 A bottling machine can be regulated so that it discharges an average of pu ounces per bottle. It has been observed that the amount of fill dispensed by the machine is Normally distributed with a 2 ounces. (a) If n 9 bottles are randomly selected from the output of the machine, what is the probability that the sample mean differs from by at most .3 ounce? (b) Find the probability that the sample mean differs from pu by at most .3 ounce when the sample size is n 25, 36, and 64, respectively. What pattern do you observe as n increascs? Can you provide an explanation for your finding? (c) How does the probability obtained in part (a) change when o is unknown, and the sample variance equals S 5.3?Explanation / Answer
a)
Note that the mean difference from the population mean (u) is 0.
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = -0.3
x2 = upper bound = 0.3
u = mean = 0
n = sample size = 9
s = standard deviation = 2
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -0.45
z2 = upper z score = (x2 - u) * sqrt(n) / s = 0.45
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.32635522
P(z < z2) = 0.67364478
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.347289559 [ANSWER]
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b)
For n = 25:
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = -0.3
x2 = upper bound = 0.3
u = mean = 0
n = sample size = 25
s = standard deviation = 2
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -0.75
z2 = upper z score = (x2 - u) * sqrt(n) / s = 0.75
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.226627352
P(z < z2) = 0.773372648
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.546745295 [ANSWER]
*******
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = -0.3
x2 = upper bound = 0.3
u = mean = 0
n = sample size = 36
s = standard deviation = 2
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -0.9
z2 = upper z score = (x2 - u) * sqrt(n) / s = 0.9
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.184060125
P(z < z2) = 0.815939875
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.631879749 [ANSWER]
******
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = -0.3
x2 = upper bound = 0.3
u = mean = 0
n = sample size = 64
s = standard deviation = 2
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -1.2
z2 = upper z score = (x2 - u) * sqrt(n) / s = 1.2
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.11506967
P(z < z2) = 0.88493033
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.76986066 [ANSWER]
************
As n increases, the probability that the sample mean differs by at most 0.3 oz from u increases as well. [ANSWER]
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c)
In that case, we use the sample standard deviation to estimate the population standard deviation.
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = -0.3
x2 = upper bound = 0.3
u = mean = 0
n = sample size = 9
s = standard deviation = sqrt(5.3) = 2.302172887
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -0.390935018
z2 = upper z score = (x2 - u) * sqrt(n) / s = 0.390935018
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.347922634
P(z < z2) = 0.652077366
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.304154731 [ANSWER]
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