516hw3 - STAT516 Solution to Homework 3 Section 2.1 4 1 6 1...

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Unformatted text preview: STAT516 Solution to Homework 3 Section 2.1 4 1 6 1 = 0.375 2. P (2 boys & 2 girls) = = 2 2 2 16 Hence, P (different number of boys & girls in a family of 4 children) = 1 - P (2 boys & 2 girls) = 1 - 0.375 = 0.625 So, in a family of 4 children, different number of boys and girls would be more common than same number of boys and girls. And the relative frequencies are: P (different number of boys & girls in a family of 4 children) = 0.625 P (same number of boys & girls in a family of 4 children) = 0.375 2 2 4. P (2 sixes in first 5 rolls | 3 sixes in 8) P (2 sixes in first 5 i 3 sixes in 8) = P (3 sixes in 8) P (2 sixes in first 5 1 six in last 3) = P (3 sixes in 8) P (2 sixes in first 5)P (1 sixes in last 3) = P (3 sixes in 8) 5 2 5 3 3 1 2 1 1 5 2 6 6 6 6 1 = 8 3 5 5 1 3 6 6 5 3 2 1 = 30 = 0.536 = 8 56 3 [as first 5 and last 3 rolls are independent] 8 4 4 ( 6. a) P (4 hits) = 0.7 ) ( 0.3) = 0.1361 4 b) P (4 hits | at least 2 hits) P (4 hits i at least 2 hits) = P (at least 2 hits) P (4 hits) = P (at least 2 hits) [as 4 hits gurantees there is at least 2 hits] 8 4 4 ( 0.7 ) ( 0.3) 4 = 8 8 1 0 8 7 1 - 0.7 ) ( 0.3) - 0.7 ) ( 0.3) ( ( 0 1 = 0.1363 c) P (4 hits | first 2 are hits) = P (2 hits in next 6 | first 2 are hits) independence 6 2 4 = P (2 hits in next 6) = 0.7 ) ( 0.3) = 0.05953 ( 2 12. a) P (makes exactly 8 bets before stopping) = P(4 wins in first 7 bets and one win on the 8th) = P(4 wins in first 7) P (win on the 8th) [independence] 7 18 20 18 = 4 38 38 38 7 5 20 3 18 = = 0.1217 4 38 38 b) P (makes at least 9 bets before stopping) = P(wins at most 4 bets in first 8) 8 8 7 8 2 20 6 8 3 20 5 8 4 20 4 20 18 20 18 18 18 = + + + + 1 2 3 4 38 38 38 38 38 38 38 38 38 4 3 = 0.6926 Section 2.2 2. Here = 204 and = 9.998 N a) P (190 x H 210) (0.65) - (-1.45) = 0.6686 N b) P (210 x H 220) (1.65) - (0.55) = 0.2417 c) P ( H =N 200) (-0.35) - (-0.45) = 0.0368 d) P ( H =N 210) (0.65) - (0.55) = 0.0333 3. a) First one. Notice that in both the cases we are calculating P( proportion of heads e 0.55). Law of large number says that with 400 trials the proportion of heads should be closer to 0.5 than in the case of only 100 trials. b) First case: = 50 and = 5 : 1 - (0.9) = 0.1814 Second case: = 200 and = 10 : 1 - (1.95) = 0.0256 6. Let X be the number of voters opposing the measure. Then X follows binomial (200, 0.45). This gives = 90 and = 7.035 for normal approximations. 90.5 89.5 - 90 - 90 90) - = a) P ( X = 0.0567 7.035 7.035 100.5 - 90 = b) P ( X >N 100) 1 - 0.0681 7.035 9. a) Think of 324 independent trials, each a "success" (the person shows up) with probability 0.9. So N = the number of arrivals, has a binomial (324, 0.9) distribution. 300.5 - 291.6 P (overbooking) = P ( N > 300) 1 - = 0.0495 5.4 b) Increase: Each group would show up with probability 0.9 approximately. So the "number of trials" gets reduced keeping the success probability same. So the distribution of proportion of success would be more flat putting more mass at the tails [larger variance as proportion 1/ n ]. c) Here N would follow binomial (162, 0.9) 150.5 - 145.8 P (overbooking) = P ( N > 150) 1 - = 0.1093 . 3.82 11. Number of hits follows binomial (100, 0.3) which gives = 30 and = 4.58 for normal approximation. a) P ( b) P ( c) P ( 31 hits) 1 - ( 0.11) = 0.4562 27 hits) 1 - ( -0.545 ) = 0.2929 33 hits) 1 - ( 0.545 ) = 0.2929 d) No: independence would be lost. This will increase the chances above. [Loss of independence would result in increased variance, so the histogram would put more mass at the tails] e) Part b) says that getting 33 or more hits has a probability of 0.2929 which is big enough to doubt that there has been a change in his batting performance. 12. 10,000 independent trials with success probability 0.5 give = 5, 000 and = 50 for normal approximation. Hence, P (5000 - m # heads 5000 + m) 5000 + m + 0.5 - 5000 5000 - m - 0.5 - 5000 - 50 50 m - + 0.5 m - 0.5 = - 50 50 m + 0.5 = 2 - 1 50 m 2 m + 0.5 + 0.5 5 = This equals if and only if = 0.97 and m = 48. , which gives 3 50 50 6 ...
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This note was uploaded on 09/14/2009 for the course STAT 134 taught by Professor Aldous during the Fall '03 term at Berkeley.

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