In the year 2014, according to salary.com, physical therapist in Jacksonville, Florida have an average annual salary of 75,000. The standard deviation is about 6,000 dollars. The distribution of salaries is slightly right skewed. Suppose that a random sample of 60 physical therapist is taken from Jacksonville, Florida. What is the sampling distribution of the sample mean salary? O Not approximately normal O approx. Normal with mean = 75.000 and stadard deviation = 6,000/sqrt(60) O approx. Normal with mean = 75,000 and stadard deviation = 6,000/(60) O approx. Normal with mean = 75,000 and stadard deviation = 6,000

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What is the sampling distribution of the sample mean salary?
In the year 2014, according to salary.com, physical therapists in Jacksonville, Florida have an average annual salary of $75,000. The standard deviation is about $6,000. The distribution of salaries is slightly right skewed. Suppose that a random sample of 60 physical therapists is taken from Jacksonville, Florida. What is the sampling distribution of the sample mean salary?

- Not approximately normal
- approx. Normal with mean = 75,000 and standard deviation = 6,000/sqrt(60)
- approx. Normal with mean = 75,000 and standard deviation = 6,000/(60)
- approx. Normal with mean = 75,000 and standard deviation = 6,000

This multiple-choice question presents four options for the distribution of the sample mean salary. Notably, the second option describes a normal distribution with a mean of 75,000 and a standard deviation given by the formula \( 6000 / \sqrt{60} \).
Transcribed Image Text:In the year 2014, according to salary.com, physical therapists in Jacksonville, Florida have an average annual salary of $75,000. The standard deviation is about $6,000. The distribution of salaries is slightly right skewed. Suppose that a random sample of 60 physical therapists is taken from Jacksonville, Florida. What is the sampling distribution of the sample mean salary? - Not approximately normal - approx. Normal with mean = 75,000 and standard deviation = 6,000/sqrt(60) - approx. Normal with mean = 75,000 and standard deviation = 6,000/(60) - approx. Normal with mean = 75,000 and standard deviation = 6,000 This multiple-choice question presents four options for the distribution of the sample mean salary. Notably, the second option describes a normal distribution with a mean of 75,000 and a standard deviation given by the formula \( 6000 / \sqrt{60} \).
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