# mids - PSTAT 5E Midterm Key Winter 2006 Question 1 Let X be...

This preview shows pages 1–3. Sign up to view the full content.

PSTAT 5E Midterm Key Winter 2006 Question 1 Let X be the number of jobs a student has (a) P (a student has 2 jobs) = P (2) = 1 – [ P (0) + P (1)] = 1 – [0.5 + 0.3] = 0.2 (b) P (a student has at least 1 job) = P (1) + P (2) = 0.3 + 0.2 = 0.5 (c) The mean number of jobs is just the E ( X ) E ( X ) = = = 3 1 i i i x p µ = 0(0.5) + 1(0.3) + 2(0.2) = 0.7 Recall that the standard deviation is ) ( X Var . So we must first find Var(X). Var(X) = = 3 1 2 ) ( i i i x p = 2 ) 7 . 0 0 ( 5 . 0 + 2 ) 7 . 0 1 ( 3 . 0 + 2 ) 7 . 0 2 ( 2 . 0 = =0.245 + 0.027 + 0.338 = 0.61 So, the standard deviation = ) ( X Var = 61 . 0 = 0.781 Question 2 (a) Let A = {odd number}, then A = {1, 3, 5}. 2 1 ) 6 1 ( 3 ) 5 ( ) 3 ( ) 1 ( ) ( = = + + = P P P A P (b) The sum of the scores is a multiple of 5 if it is equal to 5 or 10. The outcomes from throwing two dices are: 1,1 2, 13 ,1 4,1 5, 16 ,1 1,2 2,2 3,2 4, 25 , 26 ,2 1,3 2,3 3, 34 , 35 , 36 ,3 1,4 2, 43 , 44 , 45 ,4 6,4 1,5 2, 53 , 54 ,5 5,5 6,5 1,6 2, 63 ,6 4,6 5, 66 ,6 The circled outcomes correspond to the cases where the sum is a multiple of 5. Therefore, since the number of circled outcomes is 7 and the total number of outcomes is 36, P (sum is multiple of 5) = 7/36.

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

View Full Document
(c) Exactly 3 tosses required to get a Head, i.e., we need to have a Tail on the
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/02/2009 for the course PSTAT 5E taught by Professor Eduardomontoya during the Spring '08 term at UCSB.

### Page1 / 4

mids - PSTAT 5E Midterm Key Winter 2006 Question 1 Let X be...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online