{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

exam3-key - IE 230 Seat Closed book and notes 60 minutes...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
IE 230 Seat # ________ Name ___ < KEY > ___ Closed book and notes. 60 minutes. Cover page and four pages of exam. No calculators. No need to simplify answers. This test covers Section 4.6 through Chapter 6 of Montgomery and Runger, fourth edition. Remember: A statement is true only if it is always true. The random vector ( X 1 , X 2 , . . . , X k ) has a multinomial distribution with joint pmf P( X 1 = x 1 , X 2 = x 2 , . . . , X k = x k ) = x 1 ! x 2 ! . . . x k ! n ! p 1 x 1 p 2 x 2 . . . p k x k when each x i is a nonnegative integer and x 1 + x 2 + . . . + x k = n ; zero elsewhere. The linear combination Y = c 0 + c 1 X 1 + c 2 X 2 + . . . + c n X n has mean and variance E( Y ) = c 0 + i = 1 Σ n E( c i X i ) = c 0 + i = 1 Σ n c i E( X i ) and V( Y ) = i = 1 Σ n j = 1 Σ n cov( c i X i , c j X j ) = i = 1 Σ n j = 1 Σ n c i c j cov( X i , X j ) = i = 1 Σ n c i 2 V ( X i ) + 2 i = 1 Σ n 1 j = i + 1 Σ n c i c j cov( X i , X j ) . Cov( X , Y ) = E[( X −μ X ) ( Y −μ Y )] Corr( X , Y ) = Cov( X , Y ) / ( σ X σ Y ) X = Σ i = 1 n X i / n S 2 = Σ i = 1 n ( X i X ) 2 / ( n 1) Order statistics satisfy X (1) X (2) . . . X ( n ) . Score ___________________________ Exam #3, Dec 4, 2008 Schmeiser
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
IE 230 — Probability & Statistics in Engineering I Name ___ < KEY > ___ Closed book and notes. 60 minutes. 1. (3 points each) True or false. Consider two random variables X and Y with joint pdf f X , Y . (a) T F f X , Y (2,3) = f Y (3) f X | Y = 3 (2). (b) T F f X (2) = −∞ f X , Y (2, y ) dy . (c) T F E(( XY ) 2 ) = −∞ −∞ ( vw ) 2 f X , Y ( v , w ) dv dw . (d) T F f X | Y = 4 (2) = P( X = 2 | Y = 4). (e) T F f X | Y = 4 (2) = f X , Y (2,4) / f X (2). (f) T F Cov( X , X ) = Var( X ) (g) T F If X and Y are independent, then Cov( X , Y ) = 0. 2. (3 points each) Suppose that X has an exponential distribution with mean 6 minutes. Recall that the exponential distribution has the memoryless property: For all positive constants t 1 and t 2 , P( X > t 1 + t 2 | X > t 1 ) = P( X > t 2 ). For each of the following, indicate whether the expression is a constant, an event, a random variable, or undefined. (A constant has the same numerical value for every replication of the experiment.) (a) Var( X ) constant event random variable undefined (b) P( X > t 1 + t 2 | X > t 1 ) constant event random variable undefined (c) E( X | X > 6) constant event random variable undefined (d) E( X 2 ) constant event random variable undefined (e) Corr( t 1, t 2 ) constant event random variable undefined Exam #3, Dec 4, 2008 Page 1 of 4 Schmeiser
Background image of page 2
IE 230 — Probability & Statistics in Engineering I Name ___ < KEY > ___ 3. Consider the multinomial pmf (as reviewed on the front cover).
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}