First, find the average individual value.
7,300/95 = 76.8421
Z = (score-mean)/SD = (76.8421-80)/12 = ?
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population.
Find the probability that the sum of the 95 values is less than 7,300. (Round your answer to four decimal places.)
2 answers
.9876