Answer :
Answer:
Mean of sampling distribution of the sample mean alcohol consumption = 4.10
Standard deviation of the sampling distribution of the sample mean alcohol consumption = 0.14
Step-by-step explanation:
We are given that a recent sociological report states that university students drink 4.10 alcoholic drinks per week on average, with a standard deviation of 1.7505.
Suppose Jason, a policy manager at a local university, decides to take a random sample of 165 university students to survey them about their drinking habits.
Here, Population mean, [tex]\mu[/tex] = 4.10 alcoholic drinks per week
Population Standard deviation, [tex]\sigma[/tex] = 1.7505
Sample of university students = n = 165
Now, Let [tex]\bar X[/tex] = Sampling distribution of the sample mean alcohol consumption
According to Central Limit Theorem, the mean and standard deviation of the sampling distribution is given by;
Mean of sampling distribution = Population mean = [tex]\mu[/tex] = 4.10
Standard deviation of sampling distribution = Population Standard deviation ÷ [tex]\sqrt{n}[/tex]
= [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{1.7505}{\sqrt{165} }[/tex] = 0.14
Therefore, mean and standard deviation of the sampling distribution of the sample mean alcohol consumption is 4.10 and 0.14 respectively.