Assume 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. Determine the mean and standard deviation of the sampling distribution of the sample mean alcohol consumption. Provide your answer with precision to two decimal places. mean of the sampling distribution

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, = 4.10 alcoholic drinks per week

         Population Standard deviation, = 1.7505

         Sample of university students = n = 165

Now, Let = 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 = = 4.10

Standard deviation of sampling distribution = Population Standard deviation ÷

           = = = 0.14

Therefore, mean and standard deviation of the sampling distribution of the sample mean alcohol consumption is 4.10 and 0.14 respectively.

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3

answer: 270

step-by-step explanation:

you have to times 200 by 1.35 to get your answer