Answer:
the answer is D hope this helps you
Answer:
D
Step-by-step explanation:
You should use a T distribution to find the critical T value based on the level of confidence. The confidence level is often given to you directly. If not, then look for the significance level alpha and compute C = 1-alpha to get the confidence level. For instance, alpha = 0.05 means C = 1-0.05 = 0.95 = 95% confidence
Use either a table or a calculator to find the critical T value. When you find the critical value, assign it to the variable t.
Next, you'll compute the differences of each pair of values. Form a new column to keep everything organized. Sum everything in this new column to get the sum of the differences, which then you'll divide that by the sample size n to get the mean of the differences. Call this dbar (combination of d and xbar)
After that, you'll need the standard deviation of the differences. I recommend using a calculator to quickly find this. A spreadsheet program is also handy as well. Let sd be the standard deviation of the differences
The confidence interval is in the form (L, U)
L = lower bound
L = dbar - t*sd/sqrt(n)
U = upper bound
U = dbar + t*sd/sqrt(n)
The probability that the sample mean is less than 30 school days will be given as follows:
the z-score is given by:
z=(x-μ)/σ
where
μ- stand deviation
σ-mean
thus the z-score of our information is:
z=(30-28)/9
z=0.222222
hence the probability that the sample mean is less than 30 will be:
P(x<30)=0.5871~58.71%
I'm not completely sure, but I think you do 2 times itself twice which would get you 8..
