Suppose that the numbers of residents in zip codes are normally distributed with an unknown mean and standard deviation. The numbers of residents of 50 randomly sampled zip codes are used to estimate the mean of the population. What t-score should be used to find the 99% confidence interval for the population mean?

Step-by-step explanation:

The required t score would be determined from the t distribution table. In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 50 - 1 = 49

Since confidence level = 99% = 0.99, α = 1 - CL = 1 – 0.99 = 0.01

α/2 = 0.01/2 = 0.005

the area to the right of z0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995

Looking at the t distribution table,

t score = 2.678

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.

x is the total time in minutes for the free response

you are given number of questions as 15 - m and time per item as 8 minutes

to find x multiply the two givens together:

x = 8 * (15-m) = 8(15-m)