Name the appropriate distribution For each scenario below, determine the distribution you would use to answer the question. Challange: For each scenario, would you be able to find the appropriate numerical answer? Would you be able to provide a complete picture of the distribution with appropriate labels (on the axis and distribution)?a. 78% of all Millennials drink Starbucks coffee at least once a week. Suppose a random sample of 50 Millennials will be selected. What is the approximate distribution for the proportion in a sample who drink Starbucks coffee at least once a week (from all possible random samples of 50 Millennials)? a. Uniform (3, 11) b. Uniform (7,2.3) c Normal (0.78, 0.0655)d. Normal (7, 0.3637) t(39)e. Binomial(40, 0.78)f. Normal (0.78, 0.0586)g. Normal (31.2, 2.6199)

10 a) Since the sample size is 50, and we know by central limiting theorem that the distribution of proportions is normal for sample all samples with np > 5, np(1-p) > 5

Here , np = 50 x 0.78 = 39 > 5, np(1-P) = 50 x 0.78 x 0.22 = 8.58 > 5

So, the distribution of sample proportion is Normal

The mean is equal to population proportion that is 0.78, and the standard deviation is p(1-p) 0.78 x (1 – 0.78) = 0,0586 Vn TOV 50

So we get N(0.78, 0.0586) Option 7 is correct

10 b) Since the sample size is 40, and we know by central limiting theorem that the distribution of proportions is normal for sample all samples with np > 5, np(1-p) > 5

Here , np = 40 x 0.78 = 31.2 > 5, np(1-P) = 40 x 0.78 x 0.22 = 6.864 > 5

So, the distribution of sample proportion is Normal

The mean is equal to population proportion that is 0.78, and the standard deviation is p(1-p) 0.78 x (1 – 0.78) = 0,0586 Vn TOV 50

So we get N(0.78, 0.0586) Option 6 is correct

10 c) Since the waiting time is distributed uniformly between 3 to 11 minutes

So we have to use the distribution U (3,11)

Option 1st is correct

10 d) We know by central limiting theorem that the distribution of means follow normal distribution if sample size is greater than 30, and the mean of the distribution of means is equal to population mean, and the standard deviation is equal to {\sigma\over \sqrt{n}}

Here n is sample size and {\sigma } is population standard deviation

Here n = 40, y = 7,0 = 2.3

So the distribution of average waiting time is ( ) 20 170)

Option 3rd is correct

Step-by-step explanation: