4 value: 2.00 points CRA CDs Inc. wants the mean lengths of the \"cuts\" on a CD

Question

4 value: 2.00 points CRA CDs Inc. wants the mean lengths of the "cuts" on a CD to be 128 seconds (2 minutes and 8 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 6 seconds. Suppose we select a sample of 21 cuts from various CDs sold by CRA CDs Inc. a. What can we say about the shape of the distribution of the sample mean? Sample mean (Click to select) b. What is the standard error of the mean? (Round your z-value to 2 decimal places and final answer to 2 decimal places.) Standard error of the mean seconds c. What percent of the sample means will be greater than 130 seconds? (Round your z-value to 2 decimal places and final answer to 2 decimal places.) Percent d. What percent of the sample means will be greater than 123 seconds? (Round your answer to 2 decimal places.) Percent e. What percent of the sample means will be greater than 123 but less than 130 seconds? (Round your z value to 2 decimal places and final answer to 2 decimal places.)

Explanation / Answer

Solution:

Given in the question

Mean = 128 seconds

SD = 6 seconds

n= 21

Solution(a):

As we can see that population is normally distributed but sample size is less than 30 so we can not say that sample mean is normally distributed. As sample size large enough or more than 30 than we can say that sample means shape is also normally distributed.

Solution(b):

So standard error of mean = population Standard deviation/sqrt(n)

Standard error of mean = 6/(sqrt(21) = 1.3093

Solution(c):

P(Xbar>130)=1-P(Xbar<130)

Z = (130-128)/1.3093 = 2/1.3093 = 1.527

So P(Xbar>130) = 1- 0.9370 = 0.063 or 6.3%

Solution(d);

P(Xbar>123) = 1-P(Xbar<123)

Z = (123-128)/1.3093 = -5/1.3093 = -3.8188

P(Xbar>123) = 1 - 0.00007 = 0.99993 or 99.993%

Solution(e):

P(123<Xbar<130) = P(X<130) - P(X<123)

P(123<Xbar<130) = 0.9370 - 0.00007 = 0.93693 or 93.793%

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