A manufacturer of a particular brand of cereal claims the weight of a cereal in
ID: 3134664 • Letter: A
Question
A manufacturer of a particular brand of cereal claims the weight of a cereal in a box is normally distributed with a mean of 12.0 oz and a standard deviation of 0.06 oz. a. Find the probability that a randomly selected box of this cereal weighs between 11.92 oz and 12.12 oz. b. A random sample of 25 boxes was collected (before they were weighed by the company) and their weights recorded. Describe the sampling distribution of the sample mean weight of the boxes. c. What is the probability that we will find a mean weight of 11.96 or less from the sample of 25 boxes? You may use Minitab to find the standard normal probability, if you cannot find the value in the table.
Explanation / Answer
A)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 11.92
x2 = upper bound = 12.12
u = mean = 12
s = standard deviation = 0.06
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -1.333333333
z2 = upper z score = (x2 - u) / s = 2
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.09121122
P(z < z2) = 0.977249868
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.886038648 [ANSWER]
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b)
They are normally distributed with the same mean,
u(X) = 12.0 oz
and a reduced standard deviation,
sigma(X) = sigma/sqrt(n) = 0.06/sqrt(25) = 0.012 oz [ANSWER]
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c)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 11.96
u = mean = 12
n = sample size = 25
s = standard deviation = 0.06
Thus,
z = (x - u) * sqrt(n) / s = -3.333333333
Thus, using a table/technology, the left tailed area of this is
P(z < -3.333333333 ) = 0.00042906 [ANSWER]
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