CRA CDs Inc. wants the mean lengths of the \"cuts\" on a CD to be 136 seconds (2
ID: 3134731 • Letter: C
Question
CRA CDs Inc. wants the mean lengths of the "cuts" on a CD to be 136 seconds (2 minutes and 16 seconds). This will allow the disk jockeys to have plenty of time for commercials within each 10-minute segment. Assume the distribution of the length of the cuts follows the normal distribution with a population standard deviation of 5 seconds. Suppose we select a sample of 24 cuts from various CDs sold by CRA CDs Inc. What can we say about the shape of the distribution of the sample mean? What is the standard error of the mean? (Round your z-value to 2 decimal places and final answer to 2 decimal places.) What percent of the sample means will be greater than 138 seconds? (Round your z-valuo to 2 decimal places and final answer to 2 decimal places.) What percent of the sample means will be greater than 128 seconds? (Round your answer to 2 decimal places) What percent of the sample means will be greater than 128 but less than 138 seconds? (Round your z- value to 2 decimal places and final answor to 2 decimal places.)Explanation / Answer
a)
As the original distribution is normal on its own, the the sample mean distribution must be
NORMAL. [ANSWER]
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b)
SE = s/sqrt(n) = 5/sqrt(24) = 1.020620726 [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 = 138
u = mean = 136
n = sample size = 24
s = standard deviation = 5
Thus,
z = (x - u) * sqrt(n) / s = 1.96
Thus, using a table/technology, the right tailed area of this is
P(z > 1.96 ) = 0.0250 [ANSWER]
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d)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 128
u = mean = 136
n = sample size = 24
s = standard deviation = 5
Thus,
z = (x - u) * sqrt(n) / s = -7.84
Thus, using a table/technology, the right tailed area of this is
P(z > -7.84 ) = 1 [ANSWER]
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e)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 128
x2 = upper bound = 138
u = mean = 136
n = sample size = 24
s = standard deviation = 5
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -7.83
z2 = upper z score = (x2 - u) * sqrt(n) / s = 1.96
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0
P(z < z2) = 0.9750
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.9750 [ANSWER]