AP Stats Ch 6 Solutions - Chapter 6 Section 6.1 Check Your Understanding page 344 1 We are looking for the probability that the student gets either an A

AP Stats Ch 6 Solutions - Chapter 6 Section 6.1 Check Your...

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132 The Practice of Statistics for AP*, 4/e Chapter 6 Section 6.1 Check Your Understanding, page 344: 1. We are looking for the probability that the student gets either an A or a B. This probability is 0.42 0.26 0.68. + = 2. We are looking for ( ) 2 0.02 0.10 0.12. P X < = + = 3. This histogram is left skewed. This means that higher grades are more likely, but that there are a few lower grades. Check Your Understanding, page 349: 1. ( ) ( ) ( ) ( ) 0 0.3 1 0.4 2 0.2 3 0.1 1.1. X µ = + + + = The long-run average, over many Friday mornings, will be about 1.1 cars sold. 2. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 0 1.1 0.3 1 1.1 0.4 2 1.1 0.2 3 1.1 0.1 0.89. X σ = + + + = So 0.943. X σ = On average, the number of cars sold on a randomly selected Friday will differ from the mean (1.1) by 0.943 cars sold. Exercises, page 353: 6.1 (a) If you toss a coin 4 times, the sample space is {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}. All of these outcomes are equally likely and so have probability 1 0.0625. 16 = To find the probability of X taking on any specific number, count the number of outcomes with exactly this number of heads and multiply by 0.0625. For example, there are 4 ways to get exactly 3 heads so ( ) ( ) 3 4 0.0625 0.25. P X = = = This leads to the following distribution: Value 0 1 2 3 4 Probability 0.0625 0.25 0.375 0.25 0.0625
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Chapter 6: Random Variables 133 (b) The histogram shows that this distribution is symmetric with a center at 2. (c) ( ) ( ) 3 1 4 1 0.0625 0.9375. P X P X = − = = − = There is a 93.75% chance that you will get three or fewer heads on 4 tosses of a fair coin. 6.2 (a) If we roll two 6-sided dice, the sample space is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}. All of these are equally likely, so the probability of any one outcome is 1 . 36 To find the probability of X taking on any specific number, count the number of outcomes where the sum of the dice is exactly this number and multiply by 0.0278. For example, there are 4 ways to get a sum of 5 (the outcomes (1,4), (2,3), (3,2), and (4,1)) so ( ) 1 4 3 4 . 36 36 P X = = This leads to the following distribution: Value 2 3 4 5 6 7 8 9 10 11 12 Prob 1 36 2 36 3 36 4 36 5 36 6 36 5 36 4 36 3 36 2 36 1 36 (b) The histogram shows that the distribution is symmetric about a center of 7. (c) ( ) ( ) 1 2 3 6 1 5 5 1 4 1 1 1 . 36 36 36 36 6 6 P T P T = − = − + + = − = − = This means that about five-sixths of the time, when you roll a pair of 6-sided dice, you will have a sum of 5 or more.
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134 The Practice of Statistics for AP* , 4/e 6.3 (a) “At least one nonword error” is the event { } 1 X or { } 0 . X > P(X ≥1) = 1 − P(X<1) = 1 − P(X=0) = 1 − 0.1 = 0.9. (b) The event {X 2} is “no more than two nonword errors,” or “fewer than three nonword errors.” P(X 2 ) = (X = 0 ) + P(X = 1 ) + P(X = 2 ) = 0 .1 + 0.2 + 0. 3 = 0 .6. P(X < 2) = P(X = 0 ) + P(X = 1 ) = 0 .1 + 0. 2 = 0 .3. 6.4 (a) “Plays with at most two toys” is the event { } 2 X or { } 3 . X < P(X 2 ) = (X = 0 ) + P(X = 1 ) + P(X = 2 ) = 0 .03 + 0.16 + 0.30 = 0 .49. (b) The event {X > 3} is “the child plays with more than three toys.” ( ) ( ) ( ) 3 4 5 0.17 0.11 0.28. P X P X P X > = = + = = + = ( ) ( ) 3 1 2 1 0.49 0.51. P X P X = − = − = 6.5 (a) All of the probabilities are between 0 and 1 and they sum to 1 so this is a legitimate probability distribution. (b) This is a right skewed distribution with the largest amount of probability on the digit 1.
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