Achess piece is wandering around on an otherwise vacant 8×8 chessboard. at each move, the piece (a king, queen, rook, bishop, or knight) chooses uniformly at random where to go, among the legal choices (according to the rules of chess, which you should look up if you are unfamiliar with ) for each of these cases, determine whether the markov chain is irreducible, and whether it is aperiodic. hint for the knight: note that a knight’s move always goes from a light square to a dark square or vice versa. a knight’s tour is a sequence of knight moves on a chessboard such that the knight visits each square exactly once. many knight’s tours exist.(b) suppose for this part that the piece is a rook, with initial position chosen uniformly at random. find the distribution of where the rook is after n moves.(c) now suppose that the piece is a king, with initial position chosen deterministically to be the upper left corner square. determine the expected number of moves it takes him to return to that square, fully simplified, preferably in at most 140 characters.(d) the stationary distribution for the random walk of the king from the previous part is not uniform over the 64 squares of the chessboard. a recipe for modifying the chain to obtain a uniform stationary distribution is as follows. label the squares as 1, 2, . . , 64, and let di be the number of legal moves from square i. suppose the king is currently at square i. the next move of the chain is determined as follows: step 1: generate a proposal square j by picking uniformly at random among the legal moves from i. step 2: flip a coin with probability min(di/dj, 1) of heads. if the coin lands heads, go to j. otherwise, stay at i. show that this modified chain has a stationary distribution that is uniform over the 64squares.

it is probably the graph on the bottom right.

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step-by-step explanation:

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation:

just keep asking questions until you have a better understanding.

step-by-step explanation: