# selected answers for chapters 10, 11, 12 (Miller et al., ninth edition)

p.625 - 20: 16x15x14x13x12x11 = 5765760

p.625 - 32: 5^5 = 3125, since there are five people available for each day.

p.625 - 34: 12, since Bill is not there, there are only four members - 4x3=12.

p.625 - 42: 5^20 = (approx) 10^14

p.637 - 4: (13!)/(6! × 7!) = 1716

p.637 - 6: (14!)/(9!) = 14 × 13 × 12 × 11 × 10 = 240240

p.637 - 8: (13!)/(4! × 9!) = 715.

p.637 - 26: C(13, 5) = 1287.
C(26,5) = 65780.
0 (there are only four aces).

p.666 - 6: a) There are 13 possible outcomes (13 choices for the head of the line)
b) 8 (the number of girls)
c) 5 (the number of boys
d) 8/13
e) 8:5 (8 to 5)

p.666 - 8: a) {(H,H),(H,T),(T,H),(T,T)} (these are ordered pairs, the first refers to the dime, and the secon to the quarter)
b) 2/4 = 1/2 = .5
c) 2/4 = 1/2 = .5
d) 1/4 = .25
e) 2/4 = .5 (HH or TT)

p.666 - 42: a) 37:63 or 37 to 63 (.37 = 37/100, hence use 100 possible outcomes)
b) 63:37 or 63 to 37

p.675 - 4: P(not less than 2) = P(not 1) = 1 - P(1) = 1-(1/6)=5/6.

p.675 - 6: P(odd or less than 5) = P(1 or 2 or 3 or 4 or 5) = 5/6

p.675 - 16: P(11 or 12) = P((5,6) or (6,5) or (6,6)) = 3/36 = 1/12

p.684 - 2: Independent, one die does not affect the outcome of the other.

p.684 - 6: Not independent, like cards if the first selected is republican, fewer republicans remain for the second position.

p.684 - 24: 13/51; given that the first card is a diamond, only 51 remain, of which 13 are clubs.

interegnum: If A and B are independent, P(A intersect B) = .4 × .6 = .24, P(A union B) is therefore .4 + .6 - .24 = .76
If A and B are mutually exclusive, P(A intersect B) = 0, P(A union B) is therefore .4 + .6 - 0 = 1 (in this case, A and B are called complementary, we may not have used this term in this course).
If they are drawn without replacement, there are 12/52 face cards for the first card, but if the first card is a face card, only 11/51 remain for the second; hence the probability is (12 × 11)/(52 × 51)
If they are drawn with replacement (and reshuffle) there are 12/52 face cards for each draw, so the probability is (12^2)/(52^2)

p.700 - 4: .25 + .25 = .5

p.700 - 24: A) .1 × 40000 + .6 × 180000 + .3 × 250000 = 187000
B) .2 × 0 + .35 × 210000 + .45 × 290000 = 204000
C) .65 × 60000 + .35 × 340000 = 158000
Hence choose project B.

p.734 - 8: mean: 2.09/9 = .2322; median: .22 (put in rank order); midrange: (.53+.03)/2 = .28; (.28 is the mode)

p.734 - 20: 2.39 sec. (put in rank order)

p.734 - 22: 2.28 sec.

p.734 - 24: The median is affected less.(and the midrange is affected most)

p.734 - 28: (5 × 2 + 1 × 4 + 8 × 6 + 4 × 8)/(5+1+8+4) = 5.22
The median is 6; the midrange is (8+2)/2=5; the mode is 6.

p.745 - 6: The range is 84-55 = 29; s= 9.54 (mean = 70.44)

p.753 - 6: The 75th percentile is the third quartile, .75 × 40 = 30, so they would take the 31st datum or 70. However, Q3 is computed as 69.5 in the text.

p.753 - 22: The midrange gives central tendancy; I listed this problem to remind you that you are supposed to know the definition of the midrange.

p.764 - 12: This is one standard deviation or more above the mean, hence by the empirical rule, .5 - (.68/2) = .16

p.764 - 14: This is three standard deviation units above the mean, hence by the empirical rule .5 - (.997/2) = .0015