An analyst determined that the sample mean and variance for a normal distribution are 42 and 9, respectively. The 99% confidence interval for this random variable is closest to: Select exactly 1 answer(s) from the following: A. 15.0 to 69.0. B. 18.8 to 65.2. C. 34.3 to 49.7. D. 39.0 to 45.0. Thanks a lot
99% confidence interval = mean +/- 2,58SD
thanks a lot
c - for sure
so it’s B : 42-2.58*9 = 18.8; 42+2.58*9=65.2
got tricked by using variation of 9 instead of standard deviation of 3? It is C.
yeah C. read the question very very carefully, i keep telling myself. they are sneaky.