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THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION

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THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Let the random variable X follow a normal distribution with a mean of μ and a standard deviation of σ.Let THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: Let the random variable X follow a normal distribution with a mean of μ and a standard deviation of σ.Let    <sub>1</sub> be the mean of a sample of 16 observations randomly chosen from this population,and    <sub>2</sub> be the mean of a sample of 25 observations randomly chosen from the same population. -Suppose that 15% of all invoices are for amounts greater than $1,000.A random sample of 60 invoices is taken.What is the mean and standard error of the sample proportion of invoices with amounts in excess of $1,000? What is the probability that the proportion of invoices in the sample is greater than 18%?
1 be the mean of a sample of 16 observations randomly chosen from this population,and THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: Let the random variable X follow a normal distribution with a mean of μ and a standard deviation of σ.Let    <sub>1</sub> be the mean of a sample of 16 observations randomly chosen from this population,and    <sub>2</sub> be the mean of a sample of 25 observations randomly chosen from the same population. -Suppose that 15% of all invoices are for amounts greater than $1,000.A random sample of 60 invoices is taken.What is the mean and standard error of the sample proportion of invoices with amounts in excess of $1,000? What is the probability that the proportion of invoices in the sample is greater than 18%?
2 be the mean of a sample of 25 observations randomly chosen from the same population.
-Suppose that 15% of all invoices are for amounts greater than $1,000.A random sample of 60 invoices is taken.What is the mean and standard error of the sample proportion of invoices with amounts in excess of $1,000? What is the probability that the proportion of invoices in the sample is greater than 18%?

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verifed

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P= 0.15,n = 60
E( blured image
)= P = 0.1...

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