Paul plans to launch a new 3D printing business. He must understand the cost structure for existing 3D printing businesses in order be competitive. In order to plan effectively, Paul needs his estimate of monthly costs to be accurate within about $1,000 of the actual population mean. Assume that we know that there is a total population of 3,000 3D printing businesses. The mean monthly cost of doing business is $20 thousand for this population, the population standard deviation is $8 thousand. Use this information and any appropriate probability tables to solve the following problems.how many businesses should Paul survey, and why?
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
Paul plans to launch a new 3D printing business. He must understand the cost structure for existing 3D printing businesses in order be competitive. In order to plan effectively, Paul needs his estimate of monthly costs to be accurate within about $1,000 of the actual population mean. Assume that we know that there is a total population of 3,000 3D printing businesses. The mean monthly cost of doing business is $20 thousand for this population, the population standard deviation is $8 thousand. Use this information and any appropriate
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