DOC PREVIEW
UIUC STAT 400 - 400Discussion06ans

This preview shows page 1-2-3-4 out of 12 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

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 …


View Full Document

UIUC STAT 400 - 400Discussion06ans

Documents in this Course
Variance

Variance

11 pages

Midterm

Midterm

8 pages

Lecture 1

Lecture 1

17 pages

chapter 2

chapter 2

43 pages

chapter 1

chapter 1

45 pages

400Hw01

400Hw01

3 pages

Load more
Download 400Discussion06ans
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view 400Discussion06ans and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view 400Discussion06ans 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?