Acompany is placing an order for boxed lunches for a meeting of its executive board. there are 10 members of the board and each will have one lunch. the company is getting the lunches from a restaurant that has 25 varieties of boxed lunches. (a) how many different ways are there for the company to place the lunch order for the 10 lunches? note that it is not important who gets what lunch. all that matters is how many of each of the 25 possible varieties are purchased b) how many ways are there for the company to place the lunch order if the 10 lunches purchased are all different?

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.

the answer is c 13. hope you got this down : d