STAT 400 Discussion 6 Answers Spring 2015 1. The weight of fish in Lake Paradise follows a normal distribution with mean of 8.1 lbs and standard deviation of 2.5 lbs. a) What proportion of fish are between 9 lbs and 12 lbs? P ( 9 < X < 12 ) = −<<−5.21.812Z5.21.89P = P ( 0.36 < X < 1.56 ) = 0.9406 – 0.6406 = 0.3. b) Mr. Statman boasts that he once caught a fish that was just big enough to be in the top 2.5% of the fish population. How much did his fish weigh? Want x = ? such that P ( X > x ) = 0.025. P ( Z > 1.96 ) = 0.0250. x = 8.1 + 2.5 × 1.96 = 13 lbs. c) If one catches a fish from the bottom 20% of the population, the fish must be returned to the lake. What is the weight of the smallest fish that one can keep? Want x = ? such that P ( X < x ) = 0.20. P ( Z < – 0.84 ) = 0.2005 ≈ 0.20. x = 8.1 + 2.5 × ( – 0.84 ) = 6 lbs.2. Bob sells thingamabobs. His yearly salary is $27,000 plus a commission of 10% of his sales. His yearly sales are normally distributed with mean $100,000 and standard deviation $20,000. a) Find the probability that Bob earns over $40,000 in a given year. Let X denote Bob’s yearly sales. Let Y denote the amount Bob earns in a given year. Then Y = 0.10 × X + 27,000. Since X has a Normal distribution with mean $100,000 and standard deviation $20,000, Y also has a Normal distribution with mean 0.10 × 100,000 + 27,000 = $37,000 and standard deviation | 0.10 | × 20,000 = $2,000. P ( Y > 40,000 ) = −>2,000000,37000,40ZP = P ( Z > 1.50 ) = 0.0668. b) Find the missing value: With probability 67% Bob earns over _______ in a given year. Need y = ? such that P ( Y > y ) = 0.67. Find z such that P ( Z > z ) = 0.67. The area to the left is 0.33 = Φ ( z ). Using the standard normal table, z = – 0.44. x = µ + σ ⋅ z. x = 37,000 + 2,000 ⋅ ( – 0.44 ) = $36,120.3 – 4. Let X be a continuous random variable with the probability density function ( )<<−=.,,, xxcxfotherwise031 3. a) Find the value of c that makes f ( x ) a valid probability density function. Must have ()∫∞∞−dxxf = 1. 1 = ( ) ( )∫∫+−−3001 dxxcdxxc = cc 29 21+ = 5 c. ⇒ 1 = 5 c. ⇒ c = 51 = 0.20. b) Find the probability P ( X > 1 ). P ( X > 1 ) = ∫31 5dxx = 0.80.c) Find the median of the probability distribution of X. F X ( x ) = 0, x < – 1, F X ( x ) = ∫−−xdyy1 5 = 1012 x−, – 1 ≤ x < 0, F X ( x ) = ∫∫+−−xdyydyy001 5 5 = 101012 x+, 0 ≤ x < 3, F X ( x ) = 1, x ≥ 3. F X ( 0 ) = 101. ⇒ 0 < median < 3. F X ( median ) = 10median1012 + = 21. ⇒ median = 2. 4. d) Find the mean of the probability distribution of X. E ( X ) = ( )∫∞∞−⋅ dxxfx = ∫∫+−−302012 5 5 dxxdxx = 1527151+− = 1526 ≈ 1.73333. e) Find the variance of the probability distribution of X. E ( X 2 ) = ( )∫∞∞−⋅ dxxfx 2 = ∫∫+−−303013 5 5 dxxdxx = 2081201+ = 1041. Var ( X ) = E ( X 2 ) – [ E ( X ) ] 2 = 2 15261041− = 450493 ≈ 1.09556.5. Suppose S = { 2, 3, 4, 5, 6, … } and P ( k ) = ka , k = 2, 3, 4, 5, 6, … . a) Find the value of a that makes this is a valid probability model. Must have ()∑xx all P = 1. ⇒ 1 = ∑∞=2 kka = basetermfirst−1 = aa−12 . ⇒ a 2 + a – 1 = 0. ⇒ a = 251 ±−. 251 −− < 0. ⇒ a = 215 − ≈ 0.618034. Note: a = ϕ1 = ϕ – 1, where ϕ is the golden ratio. b) Find P ( outcome is divisible by 2 ). P ( outcome is divisible by 2 ) = P( 2 ) + P( 4 ) + P( 6 ) + P( 8 ) + … = a 2 + a 4 + a 6 + a 8 + … = basetermfirst−1 = 22 1 aa− = aa2 = a ≈ 0.618034.6. Suppose a discrete random variable X has the following probability distribution: f ( k ) = P ( X = k ) = ka , k = 2, 3, 4, 5, 6, … , zero otherwise. where a = ϕ – 1 ≈ 0.618034, where ϕ is the golden ratio. a) Find the moment-generating function of X, M X ( t ). For which values of t does it exist? M X ( t ) = E ( e t X ) = ∑∞=⋅2 kkktae = ∑∞=⋅2 kktea = basetermfirst−1 = tteeaa 122⋅⋅−, ⋅tea < 1 ⇔ t < ln a1 = ln ϕ ≈ 0.48121. b) Find E ( X ). M 'X ( t ) = 22222 1 1 2−−−−⋅⋅⋅⋅⋅ttttt eeeeeaaaaa = 23322 12−−⋅⋅⋅ttt eeeaaa, t < ln a1. E ( X ) = M 'X ( 0 ) = ( )232 12aaa −− = aa−−12 = 3 + a ≈ 3.618034. OR E ( X ) = ...6543265432 +++++ ⋅⋅⋅⋅⋅ aaaaa a ⋅ E ( X ) = ...54326543 ++++ ⋅⋅⋅⋅ aaaa ⇒ () 1a− ⋅ E ( X ) = ...654322 ++++++ aaaaaa = 1 2 +a. Therefore, E ( X ) = aa−+11 2 = aa−−12 = a−+111 = 3 + a ≈ 3.618034.7. A computer independently generates seven random numbers from a Uniform ( 0, 1 ) distribution. a) What is the probability that exactly three will be in the interval from ½ to 1? Let X denote the number of random numbers ( out of 7 ) that are between ½ and 1. Then X has a Binomial distribution, n = 7, p = 0.50. P ( X = 3 ) = 7 C 3 × 0.50 3 × 0.50 4 = 0.2734375. b) What is the probability that fewer than three will be in the interval from ¾ to 1? Let X denote the number of random numbers ( out of 7 ) that are between ¾ and 1. Then X has a Binomial distribution, n = 7, p = 0.25. P ( X < 3 ) = 7 C 0 × 0.25 0 × 0.75 7 + 7 C 1 × 0.25 1 × 0.75 6 + 7 C 2 × 0.25 2 …
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