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Precalculus
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Chapter 9 / Exercise 16
Precalculus
Hostetler/Larson
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Unformatted text preview: 1/31/2019 file:///Users/iuliia/Downloads/730_Assignment_3%20(1).html 730_Assignment_3 1/5 1/31/2019 730_Assignment_3 Course: Math 730-04 Professor: Dr. Greg Malen Author: Iuliia Oblasova Date:01/31/19 Assignment 3. Section 2.1. 2. Suppose that in 4-child families, each child is equally likely to be boy or girl, independently of the others. Which would then be more common: 2 boys and 2 girls, or 4-child families with different numbers of boys and girls? What would be the relative frequencies? Because sex of each child does not depend on a sex of other children, these events are independet. Also, the probability of "success" (for example, to have a boy) remain the same for each trial. Therefore, this is a binomial 1 distribution. P(boy) = P(girl) = 2 . P(have two boys) = (42) ⋅ ( 1 ) 2 2 ⋅ ( 1 2 ) 2 = - probability to have exactly two boys and, therefore, two girls. 3 8 Complement of this event is a probability to have any other numer of children is 1 − 3 8 probability to have exactly two boys and two girls is less likely than any other outcome. = 5 8 > 3 8 , so the Let x be the number of boys. x 0 P(x) 1 2 3 4 1 4 6 4 1 16 16 16 16 16 The table with relative frequencies to have a certain number of boys is represented above. It could be observed that the most likely outcome of 2 boys and 2 girls is less than the sum of other outcomes 1 16 + 4 16 + 4 16 + 1 16 = 5 8 . 6. A man fires 8 shots at a target. Assume that the shots are independent, and each shot hits the bull eye with probability 0.7? Similar to the previous exercise, because events are independent and a probability of a success (to shot a target) = 0.7 remains the same for each trial, number of times a man fires a target has a binomial distribution. a). What is the chance that he hits the bull's eye exactly 4 times? P(exactly 4) = (8) ⋅ (0.7) 4 4 ⋅ (0.3) 4 ≈ 0.136 . b). Given that he hit the bull's eye at least twice, what is the chance that he hit the bull's eye exactly 4 times? P(exactly 4| at least 2) = (exactly4∩atleast2) P(atleast2) = P(exactly4) 1−P(lessthan2) 0.136 = 1−[ 8 0 8 ⋅0.7 ⋅0.3 + (0) ≈ 0.136. 8 1 7 ⋅0.7 ⋅0.3 ] (1) c). Given that the first two shots hit the bull's eye, what is the chance that he hits the bull's eye exactly 4 times in the 8 shots? In other words, what is the probability that a man will fire a target exactly 2 times out of the remaining 6. P(A) = (62) ⋅ 0.72 ⋅ 0.3 4 ≈ 0.0595 . 8. For each positive integer n, what is the largest value of p such that zero is the most likely number of successes in n independent trials with success probability of p? Known that mode (i.e. the most likely number of successes) of the binomial distribution is ⌊np+p⌋ is 0, we assume that np+p <1 by the definition of the floor function. Then, 0≤ p(n+1)<1.Therefore, p < file:///Users/iuliia/Downloads/730_Assignment_3%20(1).html 1 n+1 . 2/5 1/31/2019 730_Assignment_3 12. A gambler decides to keep betting on red at roulette, and stop as soon as the has won a total of five bets? a). What is the probability that she has to make exactly 8 bets before stopping? To make exactly 8 bets means to win 4 times out of 7 first attempts and then win the 8th bet. Assuming that the roulette has 18 red slots and 38 slots total, P(win) = 7 (4) ⋅ ( 9 19 ) 4 ⋅ ( 10 19 ) 3 9 ⋅ 9 19 . Therefore, P(have exactly 8 bets) = ≈ 0.122. 19 b). What is the probability that she has to make at least 9 shots? In other words, what is the probability that a gambler will win 4 games at maximum during first 8 games. P(4 4 wins among 8 games) = ∑i=0 8 (i ) ⋅ ( 9 19 ) i ⋅ ( 10 19 ) 8−i ≈ 0.693. Section 2.2. 2. Recalculate the approximation above for a biased coin with P(heads) = 0.51. a.)Given that n=400, μ=400⋅0.51=204 and σ = (400 ⋅ 0.51 ⋅ 0.49) 0.5 ≈ 9.998. Then, P(190 ≤ H ≤ 210) = P( 190− 1 2 −204 9.998 210+ ) ≤ z ≤ 1 2 −204 ) = P(−1.45 ≤ z ≤ 0.65) ≈ ϕ(−1.45) − ϕ(0.65) ≈ 1 − 0.4265 − 0.2422 ≈ 0. 9.998 b). P(210 ≤ H ≤ 220) = P( 210− 1 2 −204 9.998 220+ ) ≤ z ≤ c). P(H = 200) = P( d). P(H = 210) = P( 1 2 −204) ) = P(0.55 ≤ z ≤ 1.65) ≈ ϕ(1.65) − ϕ(0.55) ≈ 0.4505 − 0.2088 ≈ 0.2417 9.998 200− 1 2 −204 9.998 210− 1 2 200+ ) ≤ z ≤ −204 2 −204 ) = P(−0.45 ≤ z ≤ −0.35) ≈ 0.1736 − 0.1368 ≈ 0.0368. 9.998 210+ ) ≤ z ≤ 9.998 1 1 2 . −204 ) = P(0.55 ≤ z ≤ 0.65) ≈ 0.2422 − 0.2088 ≈ 0.0334. 9.998 9. Airline overbooking. An airline knows that over the long run, 90% of passengers who reserve seats show up for their flight. On a particular flight with 300 seats, the airline accepts 324 reservations. a). Assuming that passengers show up independently of each other, what is the chance that the flight will be overbooked? Let's define X as the number of passengers who show up on their flight and p = 0.9 - probability of success. Then, μ = 0.9 ⋅ 324 = 291.6 and σ = (324 ⋅ 0.9 ⋅ 0.1) 0.5 = 5.4. The flight will be overbooked if (300 < X≤ 324). P(300 < X≤ 324) = P( 300+ 1 2 −291.6 5.4 324+ < z ≤ 1 2 −291.6 5.4 ) = P(1.65 < z ≤ 6.09) ≈ P(z > 1.65) ≈ 0.5 - ϕ (1.65) ≈ 0.0495. b). Suppose that people tend to travel in groups. Would that increase or decrease the probability of overbooking? Explain. The probability will be higher because the standart deviation is higher, which means that events which lie within a few standard deviations from the mean are more likely to occur. c). Redo the calculation. a). assuming that passengers always travel in pairs. Instead of 324 independent passengers we have 162 independent couples. The flight will be overbooked if the number of couples who show up for the flight will be greater than 150 with μ = 162⋅0.9 = 145.8 and σ = 3.81. P(X>150) = P(z > 150+ 1 2 −145.8 3.81 ≈ P(z > 1.23) ≈ 0.5 − 0.3907 ≈ 10.93 , so indeed the probability of a flight of being overbooked is more than two times higher than in case when all passenger travel independently. 10. A probability class has 30 students. As part of an assignment, each student tosses a coin 200 times and records the number of heads. Approximately what is the chance that no student gets exactly 100 heads? file:///Users/iuliia/Downloads/730_Assignment_3%20(1).html 3/5 1/31/2019 730_Assignment_3 = 200⋅ 0.5 = 100 and σ = (200 ⋅ 0.5 ⋅ 0.5) to have exactly a 100 heads. P(X = 100) = μ 100− P(100 ≤ X ≤ 100) = P( 1 −100 2 0.5 100+ ≤ z ≤ 7.071 ≈ 7.071 1 . First, let's calculate a probability for a single student −100 2 = P(−0.071 ≤ z ≤ 0.071) = 2 ⋅ P(z ≤ 0.71) ≈ 2 ⋅ 0.0 7.071 Let's use a binomial distribution to calculate the probability not to have exactly 100 heads for any of the 0 3 students. P(noone gets a 100 heads) = (30 (0.0558) (0.9442) 0 0 ) ≈ 0.1786 . Let's use a Poisson distribution to approximate the same result: μ = 30 ⋅ 0.0568 ≈ 1.674. P(X=0)=e−1.674 ⋅ 1.674 0! 0 ≈ 0.1875 , which is fairly close. 12. A fair coin is tossed 10000 times. Find a number m such that the chance of the number of heads being between 5000-m and 5000+m is approximately 2/3? μ = 5000 and σ = 50. P(5000-m ≤ X ≤ 5000 + m) = ϕ( 5000+m+0.5−5000 m ≈ m+0.5 50 ) − ϕ( 50 0.667+1 2 5000−m−0.5−5000 ) = ϕ( 50 m+0.5 50 ) − ϕ( −m−0.5 50 ) = 2ϕ( m+0.5 50 − 1) ≈ 0.667 . Therefore, . Then, ≈ 0.96 and m ≈ 47.5 ≈ 48. Section 2.3. 1. Suppose you knew the consecutive odds ratios R(k) = P(k)/P(k-1) of a distribution P(0),...,P(n). Find a formula for P(k) in terms of R(1),...,R(n). Thus the consecutive odds ratios determine a distribution. 2. A fair coin is tossed 10000 times. The probability of getting exactly 5000 times heads is closest to which number? μ = 5000, σ = 50. P(5000 ≤ x ≤ 5000) = P( 5000−0.5−5000 50 ≤ z 5000+0.5−5000 50 ) = P(−0.01 ≤ z ≤ 0.01) = 2P(z ≤ 0.01) ≈ 2 ⋅ 0.0 Therefore, the probability of this event is closest to a value of 0.01. Additional problem. 11. Consider a biased coin with P[heads] = .51. Compute the probability of observing more heads than tails for N tosses. a). Calculate the exact probability for N = 3 tosses. More heads than tails means 3 heads and no tails or 2 heads and 1 tail. Using binomial distribution, we have 3 3 (3) ⋅ 0.51 0 ⋅ 0.49 3 2 + ( ) ⋅ 0.51 2 1 ⋅ 0.49 ≈ 0.514998 . b). Calculate the probability using the normal approximation for N = 3 tosses. μ = 1.53, σ = (3 ⋅ 0.51 ⋅ 0.49) 0.5 ≈ 0.866. Let X be the number of heads. Then P(X≥ 2) = P(z ≥ 2−1.53−0.5 0.866 ) = P(z ≥ −0.003) = 0.5 + P(0 ≤ z ≤ 0.003) ≈ 0.5 + 0.012 ≈ 0.512 , which is fairly close to the results we recieved in part a). file:///Users/iuliia/Downloads/730_Assignment_3%20(1).html 4/5 1/31/2019 730_Assignment_3 c). Calculate the probability using the normal approximation for N = 100 tosses. For N=100 tosses, in odred to have more heads than tails X has to be ≥ 51. In this case, μ = 100 ⋅ 0.51 = 51, σ = (100 ⋅ 0.51 ⋅ 0.49) 0.5 ≈ 4.999. Then P(X≥ 51 ) = P(z ≥ 51−51−0.5 4.999 ) = . P(z ≥ −0.1) = P(0 ≤ z ≤ 0.1) + 0.5 ≈ 0.5 + 0.0398 ≈ 0.5398 d). What happens to this probability as N goes to infinity? According to the law of large numbers, as N approaches infinity the results should be close to the expected file:///Users/iuliia/Downloads/730_Assignment_3%20(1).html 5/5 ...
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