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Answer:
The mean of the sampling distribution of the sample mean alcohol consumption is 4.10 alcoholic drinks per week on average.
The standard deviation of the sampling distribution of the sample mean alcohol consumption is 0.17.
Step-by-step explanation:
We have these following informations:
Assume a recent sociological report states that university students drink 4.10 alcoholic drinks per week on average, with a standard deviation of 1.9101.
Determine the mean and standard deviation of the sampling distribution of the sample mean alcohol consumption.
The mean of the sampling distribution of the sample mean is the same of that university students. So the mean of the sampling distribution of the sample mean alcohol consumption is 4.10 alcoholic drinks per week on average.
The standard deviation of the sampling distribution of the sample mean alcohol consumption is the division of the standard deviation of the sample mean divided by the length of the sample. So
[tex]s = \frac{1.9101}{\sqrt{125}} = 0.17[/tex]
The standard deviation of the sampling distribution of the sample mean alcohol consumption is 0.17.
The mean and standard deviation of sampling distribution is 4.10 and 0.17 respectively
What is z score?
Z score is used to determine by how many standard deviations the raw score is above or below the mean.
It is given by:
z = (raw score - mean) / standard deviation
Mean = 4.10, standard deviation = 1.9101, sample = 125 students. hence:
Mean of sampling distribution = 4.10
standard deviation of sampling distribution = 1.9101 / √125 = 0.17
The mean and standard deviation of sampling distribution is 4.10 and 0.17 respectively
Find out more on z score at: https://brainly.com/question/25638875
