Alist of n items is arranged in random order; to find a requested item, they are searched sequentially until the desired item is found. what is the expected number of items that must be searched through, assuming that each item is equally likely to be the one requested? (questions of this nature arise in the design of computer algorithms.)

n/2 = average number of items to search. Or more precisely (n+1)/2 I could just assert that the answer is n/2, but instead I'll prove it. Since each item has the same probability of being searched for, I'll simulate performing n searches on a list of n items and then calculate the average length of the searches. So I'll have 1 search with a length of 1, another search looks at 2, next search is 3, and so forth and so on until I have the nth search looking at n items. The total number of items looked at for those n searches will be: 1 + 2 + 3 + 4 + ... + n Now if you want to find the sum of numbers from 1 to n, the formula turns out to be n(n+1)/2 And of course, the average will be that sum divided by n. So we have (n(n+1)/2)/n = (n+1)/2 = n/2 + 1/2 Most people will ignore that constant figure of 1/2 and simply say that if you're doing a linear search of an unsorted list, on average, you'll have to look at half of the list.

False according to edgenuity