Ten people at a party decide to compare ages. Five people are 30 years old, three are 32, one is 31, and one is 65. Given the ages of the ten people, to the nearest tenth, determine the average of their ages and the standard deviation. A. The mean age of the ten people at a party is B. The standard deviation is: NEXT QUESTION TURN IT IN ASK FOR HELP 3 DLL​

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .

you divide them and then you get the answer i took this class this year

step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.