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Unformatted text preview: Stat 20: solutions to homework 7 Michael Lugo October 25, 2010 1 Chapter 18, # 2 (a) The average of the box is (1+3+5+7) / 4 = 4, and the SE is q (3 2 + 1 2 + 1 2 + 3 2 ) / 4 = √ 5 Therefore the sum of 400 draws has expected value 4 × 400 = 1600 and SE √ 400 × √ 5 = 20 √ 5 ≈ 44 . 7. 1500 is thus (1500 1600) / 44 . 7 ≈  2 . 25 SE from the average. The probability of getting at least 1500 from the sum of draws is (1 + 0 . 9756) / 2 or about 99%. (b) The expected number of 3s is 100, with standard error (1 0) q (1 / 4)(3 / 4) × √ 400 = 5 √ 3 ≈ 8 . 7. Thus 90 is about 1 . 15 SD below the average. The prob ability of having fewer than 90 3s is therefore about (1 . 7499) / 2 or about 13%. 2 Chapter 18, # 4 The expected number of heads is 25 / 2 = 12 . 5, and the SE of the number of heads is q 25 × 1 / 2 × 1 / 2 = 2 . 5. We want the area under the normal curve between 11 . 5 and 12 . 5. In standard units this is the area between (11 . 5 12 . 5) / 2 . 5 = . 4 and (12...
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This note was uploaded on 07/21/2011 for the course STAT 20 taught by Professor Staff during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Staff
 Statistics, Probability

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