quiz5sol

# quiz5sol - Statistics 21 Quiz 5 Solutions Due 1 Box has 38...

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Unformatted text preview: Statistics 21: Quiz 5 Solutions Due: April 20, 2009 1. Box has 38 tickets: 1 ticket with \$35 on it and 37 tickets with \$-1 on them. EV = Average of Box × number of draws = 1(35)- 37(- 1) 38 × 100 = \$- 5 . 26. SE = SD of Box × √ number of draws = (35-- 1) q 1 38 × 37 38 × √ 100 = \$57 . 63 . EV ± 2SE = (- 120 . 52 , 110). The most you can possibly lose in 100 draws is \$100, so the Normal approximation is not going to be a great approximation. If you use the normal approximation, you get z = 70-- 5 . 26 57 . 63 = 1 . 3. Using the normal curve, the area to the right of 1.3 is 100- 80 . 64 2 = 9 . 68%. To compute the area exactly, notice that you get over \$70 if you win 5 times or more. Then P (win 5 times or more) =1- P (win 0-4 times) =1- 100 1 38 37 38 100- 100 1 1 38 1 37 38 99- 100 2 1 38 2 37 38 98- 100 3 1 38 3 37 38 97- 100 4 1 38 4 37 38 96 = . 1244 or about 12.4%. 2. Our estimate of the percent of people who have never cheated on a test is 250 300 × 100% = 83 . 3%....
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## This note was uploaded on 07/08/2010 for the course STATS 21 taught by Professor Ibser during the Spring '09 term at Berkeley.

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