Answer:
4.10
0.17
Step-by-step explanation:
From the question, we know that
A recent sociological report stated that university students drink 4.10 alcoholic drinks per week on average, with a standard deviation of 1.9101.
Also, we're tasked with finding the mean and standard deviation of the sampling distribution of the sample mean alcohol consumption.
To find the mean of the sampling distribution, we take the sample mean of the same of that university students as a direct substitute. This then means that the mean of the sampling distribution of the sample mean alcohol consumption is 4.10 alcoholic drinks per week on average.
On the other hand, the standard deviation of the sampling distribution of the sample mean alcohol consumption is taken to be the division of the standard deviation of the sample mean. Mathematically, we have
Standard Deviation, S = 1.9101 / √125
Standard Deviation, S = 1.9101 / 11.18
Standard Deviation, S = 0.17
Therefore, the Standard Deviation is 0.17
Answer:
The values are
[tex]\mu_{\= x } = \mu = 4.10[/tex]
And
[tex]\sigma _{\= x} = 0.38202[/tex]
Step-by-step explanation:
The population mean is [tex]\mu = 4.10[/tex]
The standard deviation is [tex]\sigma = 1.9101[/tex]
The sample size is n = 125 students
Generally the mean of the sampling distribution of the sample mean is equivalent to the population mean
i.e [tex]\mu_{\= x } = \mu = 4.10[/tex]
Generally the standard deviation of the sampling distribution of the sample mean alcohol consumption is mathematically represented as
[tex]\sigma _{\= x} = \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]\sigma _{\= x} = \frac{1.9101 }{\sqrt{125} }[/tex]
=> [tex]\sigma _{\= x} = 0.38202[/tex]
