Answer:
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
Answer:
95.15%
Step-by-step explanation:
We have that the mean (m) is equal to 20, the standard deviation (sd) 3
They ask us for P (x <25)
For this, the first thing is to calculate z, which is given by the following equation:
z = (x - m) / sd
We have all these values, replacing we have:
z = (25 - 20) / (3)
z = 1.66
With the normal distribution table (attached), we have that at that value, the probability is:
P (z <1.66) = 0.9515
Which means that the probability that it arrives before 25 minutes is 95.15%
