The hours a week people spend answering and sending emails is normally distributed with a standard deviation of Ï = 150 minutes. How large of a sample size would you need to estimate the mean number of minutes people spend answering and sending emails with an error of 1 minutes with 99% confidence? What would be the practical problem with attempting to find this confidence interval?

sample size n would be 149305 large

Value of n (149305) is too high, this will be the practical problem with attempting to find this confidence interval

Step-by-step explanation:

Given that;

standard deviation α = 150 min

confidence interval = 99%

since; p( -2.576 < z < 2.576) = 0.99

so z-value for 99% CI is 2.576

E = 1 minutes

Therefore

n = [(z × α) / E ]²

so we substitute

n = [(2.576 × 150) / 1 ]²

n = [ 386.4 ]²

n = 149304.96 ≈ 149305

Therefore sample size would be 149305 large

Value of n is too high, that would be the practical problem with attempting to find this confidence interval

the answer is 8

step-by-step explanation:

3.14 x 8x8

f(x) = y

f(-3) = -5

step-by-step explanation: