Assume a recent sociological report states that university students drink 5.10 alcoholic drinks per week on average, with a standard deviation of 1.3401. Suppose Jason, a policy manager at a local university, decides to take a random sample of 150 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 sampling distribution = 5.10 alcoholic drinks per week

Standard deviation of the sampling distribution = 0.11

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 5.10 alcoholic drinks per week

Standard Deviation, σ = 1.3401

Sample size, n = 150

a) Mean of sampling distribution

The best approximator for the mean of the sampling distribution is the population mean itself.

Thus, we can write:

b) Standard deviation of the sampling distribution

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

answer:   b

step-by-step explanation:

x2 = 16

x can not be a negative number because 16 is positive

the only answer would be positive 8