Answer :
Answer:
a) Sample mean = 4.10 alcoholic drinks per week
b) Sample standard deviation = 0.14
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 4.10
Sample size, n = 125
Population standard deviation, σ = 1.7505
We have to estimate the following:
a) Sample mean alcohol consumption
[tex]\bar{x} = \mu = 4.10[/tex]
b) Standard deviation of the sampling distribution of the sample mean alcohol
[tex]s = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1.7505}{\sqrt{165}} =0.13627 \approx 0.14[/tex]
Thus, the sample mean is 4.10 alcoholic drinks per week and sample standard deviation is 0.14 alcoholic drinks per week.