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1.X x 1 2 3 4 5 6 f(x) 0.1 0.15 0.2 K 0.25 0.1 :(1) K=? [K=0.2] (2)[3.65] (3)[0.75] (4)[0.4] (5)[2.23] (6) Y=4+2X E(Y)=[11.3]
1.X
| x | 1 | 2 | 3 | 4 | 5 | 6 |
| f(x) | 0.1 | 0.15 | 0.2 | K | 0.25 | 0.1 |
:(1) K=? [K=0.2] (2)[3.65] (3)[0.75]
(4)[0.4] (5)[2.23]
(6) Y=4+2X E(Y)=[11.3] Var(Y)=[8.92]
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Auditing Cases An Interactive Learning Approach
Authors: Steven M Glover, Douglas F Prawitt
4th Edition
0132423502, 978-0132423502
