Homework 4 (100pt) Due in class Friday 9/21
Transcription
Homework 4 (100pt) Due in class Friday 9/21
Homework 4 (100pt) Due in class Friday 9/21 1. (8) 1.59 (page55) X~Binomail(1000, 1/200) Pr ( X 8) 1 Pr ( X 7) 1 0.8672 0.133 Grading note: some might use Poisson approximation and get an answer of 0.131 Pr (5 X 10) Pr ( X 5,6,7,8,9,10) 0.546 Grading note: some might use Poisson approximation and get an answer of 0.544 2. (8) 1.67 (page57) (a) The area under the density must equal 1. To verify this, make the substitution u = x2/(22), which gives du = xdx/2 , so 0 x 2 e x 2 /(2 2 ) dx = e u du = e u 0 upper limit x = c, the integral becomes: Proportion(x c) = c 0 1 - ec 2 /(2 2 ) 0 x 2 = 0 - (-e0) = 1. For a finite e x 2 /(2 2 ) dx = e u c 2 /(2 2 ) 0 . (b) For = 100, Proportion(x 200) = 1 - e 200 /( 2(100) 2 2 ) = 1 - e-2 = 1 - .135335 = .8647. Proportion(x 200) = 1 - Proportion(x < 200) = 1 - .8647 = .1353. Proportion(100 x 200) = Proportion(x 200) - Proportion(x 100) = .8647 - [1- e 100 /( 2(100) ) ] = .8647 - [1-e-.5 ] = .4712. 2 2 3. (8) 1.73 (page 58) Pr(a randomly selected bottle has a bursting strength > 300 psi)=Pr(X>300)=Pr(Z>1.67)=10.9525=0.0475 Define Y as the number of bottle in a carton that have a bursting strength >300 , Y~Binomial(12, 0.0475) 12 0 12 0.0475 (1 0.0475) =1-0.5577=0.4423 0 Pr(Y 1)= 1- Pr(Y=0)=1- = 4. (16) Case study. Determine the PMF for the following random variables. Identify which one(s) of the random variable is Binomial. a) Of all customers purchasing automatic garage-door openers, 75% purchase a chain-driven model. Let X be the number among the next 15 purchaser who select the chain-driven model Define the PMF of X and decide whether X is Binomially distributed. 15 x 15 x 0.75 (1 0.75) x X ~ Binomial(15, 0.75) , the PMF of X: Pr(X=x)= b) A pediatrician wishes to recruit 5 couples, each of whom is expecting their first child, to participate in a new natural childbirth regimen. Let p 0.2 be the probability that a randomly selected couple agrees to participate. Define X be the number of patients the pediatrician asks before she finds 5 couples who agree to participate. (Hint: consider the case where she needs only 1 couple, what is the probability that she asks 1 couple to find 1 couple who agree, how about the probability that she ask 2 couples to find 1 couple? then consider the case where she needs 2 couples, what is the probability that she asks 2 couples to find 2 agree, how about asks 3 couples to find 2 agree… until you find the pattern) X is not Binomial because sample size n is not fixed. x 1 x 5 5 (1 0.2) (0.2) 5 1 The PMF of X: Pr(X=x)= This is because for a total of x couples she asks, 5 couples agree , hence (0.2)5 , x 5 couples do not agree , hence (1 0.2) x5 . Also for a total of x couples, the last one must agree, and the other 4 x 1 5 1 who agree can be arranged in the first x 1 , hence Define the PMF of X and decide whether X is Binomially distributed. c) Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerators are running. Suppose that 7 of these refrigerators have a defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order, let X be the number among the first 6 examined that have a defective compressor. (Hint: for example, the total number of ways to arrange 7 defective 12 . Suppose there are 2 defective happen in the first 6 examined, 7 refrigerators out of 12 is 6 2 these 2 can happen in a total of ways; then the second 6 must contain 7-2=5 defectives, 6 5 these 5 defectives can happen in a total number of arrangements, what is the probability that 2 defectives are arranged to happen in the first 6 examined? ) Define the PMF of X and decide whether X is Binomially distributed. X is not Binomial because p is not fixed. 6 6 x7 x The PMF of X is Pr(X=x)= , where x 1, 2,3, 4,5,6 12 7 5. (10) 1. 30 (page 41) (a) Proportion(z 2.15) = .9842 (Table I). Proportion(z < 2.15) will also equal .9842 because the z distribution is continuous. (b) Using the symmetry of the z density, Proportion(z >1.50) = Proportion(z < -1.50) = .0668. Proportion(z > -2.00) = 1- Proportion(z -2.00) = 1 - .0228 = .9772. (c) Proportion(-1.23 z 2.85) = Proportion(z 2.85) - Proportion(z -1.23) = .9978 - .1093 =.8885. (d) In Table I, z values range from -3.8 to +3.8. For z < -3.8, the left tail areas (i.e., table values) are .0000 (to 4 decimal places); for z > +3.8, left tail areas equal 1.0000 (to 4 places). Therefore, Proportion(z > 5) = 1 - Proportion(z 5) - 1 - 1.0000 = .0000. Similarly, Proportion(z >-5) = 1 - Proportion(z -5) = 1 - .0000 = 1.0000. (e) Proportion(z < 2.50) = Proportion( -2.50 < z < 2.50) = Proportion(z < 2.50) - Proportion(z < -2.50) = .9938 - .0062 = .9876. 6. (10) 1.32 (page 41) (a) Proportion(z z*) = .9082 when z* = 1.33 (Table I). (b) Proportion(z 1.33) = .9082 and Proportion(z 1.32) = .9066; using linear interpolation, .9080 is (.9080-.9066)/(.9082-.9066) = .0014/.0016 = .875 (or, 87.5%) of the way from .9066 to .9082, so z* is .875 of the way from 1.32 to 1.33; i.e., z* = 1.32 + .875(1.33-1.32) = 1.32875, or, about 1.329. (c) Proportion(z > z*) = .1210 is a right-tail area. Converting to a left-tail area (so that we can use table I), Proportion(z z*) = 1 - .1210 = .8790. From Table I, z* = 1.17 has a left tail area of .8790 and, therefore, has a right-tail area of .1210. (b) Proportion(-z* z z*) = .754. Because the z density is symmetric the two tail areas associated with z < - z* and z > z* must be equal and they also account for all the remaining area under the z density curve. That is, 2Proportion(z - z*) = 1 - .754 = .246, which means that Proportion(z - z*) = .1230. From Table I, Proportion(z -1.16) = .1230, so - z* = -1.16 and z* = 1.16. (c) Proportion(z > z*) = .002 is equivalent to saying that Proportion(z z*) = .998. From Table I, Proportion(z 2.88) = .9980, so z* = 2.88. Similarly, you would have to go a distance of 2.88 (i.e., 2.88 units to the left of 0) to capture a left-tail area of .002. 7. (10) 1.38 (page 42) (a) When standardized, x > 100 becomes z > (100-96)/14 or, z > .29. From Table I, Proportion(z > .29) = 1 - Proportion(z .29) = 1 - .6141 = .3859. (b) When standardized, x 75 becomes z (75-96)/14 or, z > -1.5. From Table I, Proportion( z -1.5) = 1 - Proportion(z < -1.5) = 1 - .0668 = .9332. (b) When standardized, 50 < x < 75 becomes (50-96)/14 < z < (75-96)/14 or, -3.29 < z < -1.5. From Table I, Proportion(-3.29 < z < -1.5) = Proportion(z < -1.5) - Proportion(z < -3.29) = .0668 - .0005 = .0663. (c) Proportion(x a) = .05, when standardized, becomes Proportion(z a 96 14 ) = .05. From Table I, 5% of the area under the z curve lies to the left of z = -1.645. Therefore, a = 96 -1.645(14) = 72.97. For the right-tail area, 119.03. b 96 14 a 96 14 = -1.645, so = +1.645, so b = 96+1.645(14) = 8. (12) Suppose that the reaction time (sec) to a certain stimulus is a continuous random variable with a density function given by f ( x) { k / x 1 x 10 0 otherwise a) Find the numerical value of k 10 1 k / x 1 k= 1/ln10=0.434 b) Find the mean and std of x 10 E ( X ) xk / xdx k (10 1) 9k 9 / ln10 =3.906 or 3.91 1 X E ( X 2 ) ( EX )2 = 49.5k (9k )2 = 6.226 =2.495 or 2.5 c) Find the mean of 2x E(2 X ) 2*3.906 7.812 d) Find the mean of x 2 . 10 E ( X 2 ) x 2 k / xdx = 49.5k 21.483 1 9. (8) Use normal approximation on Binomial to compute the following questions with continuity correction , suppose X ~ Binomial (500,0.15) a). Pr ( X 400) X ~ Normal (500(0.15), 500(0.15)(1 0.15)) or X ~ Normal (75,7.984) Pr ( X 400) Pr ( X 399) Pr ( X 399.5) 1 b). Pr (200 X 450) Pr (200 X 450) Pr (200.5 X 450.5) 0 10. (10) Discussion question. Discuss on-line via Blackboard Learn. Discuss only, no need to turn in your work, grading will be based on your level of participation. a). Suppose X is the cost of an appetizer and Y be the cost of a main course at a certain restaurant . Suppose a customer always order both courses. The price and probability of an order is given below: Note Pr ( X 5, Y 10) 0.2 means the probability of order a $5 appetizer AND a $10 main course is 20%. Pr ( X Pr ( X Pr ( X Pr ( X Pr ( X Pr ( X Pr ( X Pr ( X Pr ( X 5, Y 10) 0.2 5, Y 15) 0.15 5, Y 20) 0.05 6, Y 10) 0.1 6, Y 15) 0.15 6, Y 20) 0.1 7, Y 10) 0.1 7, Y 15) 0.1 7, Y 20) 0.05 Do you think the order (cost) of appetizer and main course are independent? b). Suppose X ~ Normal ( , ) , Y ~ Normal ( , ) , What is E(X+X), Var(X+X), E(X+Y) and Var(X+Y).