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IPS6e.ISM.Ch05

# IPS6e.ISM.Ch05 - Chapter 5 Solutions 5.1(a n = 1500(the...

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Chapter 5 Solutions 5.1. (a) n = 1500 (the sample size). (b) The “Yes” count seems like the most reasonable choice, but either count is defensible. (c) X = 525 (or X = 975). (d) ˆ p = 525 1500 = 0 . 35 (or ˆ p = 975 1500 = 0 . 65). 5.2. n = 200 (the sample size), ˆ p = 40% = 0 . 40, and X = n ˆ p = 80. 5.3. We have 20 independent trials, each with probability of success (heads) equal to 0.5, so X has a B ( 20 , 0 . 5 ) distribution. 5.4. Assuming no multiple births (twins, triplets, quadruplets), we have four independent trials, each with probability of success (type O blood) equal to 0.25, so the number of children with type O blood has a B ( 4 , 0 . 25 ) distribution. 5.5. (a) For a B ( 4 , 0 . 3 ) distribution, P ( X = 0 ) = 0 . 2401 and P ( X 3 ) = 0 . 0837. (b) For a B ( 4 , 0 . 7 ) distribution, P ( X = 4 ) = 0 . 2401 and P ( X 1 ) = 0 . 0837. (c) The number of “failures” in a B ( 4 , 0 . 3 ) distribution has a B ( 4 , 0 . 7 ) distribution. With 4 trials, 0 successes is equivalent to 4 failures, and 3 or more successes is equivalent to 1 or fewer failures. 5.6. (a) For a B ( 100 , 0 . 5 ) distribution, µ ˆ p = p = 0 . 5 and σ ˆ p = p ( 1 p ) n = 1 20 = 0 . 05. (b) No; the mean and standard deviation of the sample count are both 100 times bigger. (That is, ˆ p = X / 100, so µ ˆ p = µ X / 100 and σ ˆ p = σ X / 100.) 5.7. (a) ˆ p has approximately a Normal distribution with mean 0.5 and standard deviation 0.025, so P ( 0 . 4 < ˆ p < 0 . 6 ) = P ( 2 < Z < 2 ) . = 0 . 9544. (b) P ( 0 . 45 < ˆ p < 0 . 55 ) = P ( 1 < Z < 1 ) . = 0 . 6826. Note: The answers given are from Table A or software. The 68–95–99.7 rule would give the answers 0.95 and 0.68—quite reasonable answers, especially since this is an approximation anyway. For comparison, the exact answers are P ( 0 . 4 < ˆp < 0 . 6 ) . = 0 . 9431 or P ( 0 . 4 ˆp 0 . 6 ) . = 0 . 9648, and P ( 0 . 45 < ˆp < 0 . 55 ) . = 0 . 6318 or P ( 0 . 45 ˆp 0 . 55 ) . = 0 . 7287. (Notice that the “correct” answer depends on our understanding of “between.”) 5.8. (a) P ( X 3 ) = ( 4 3 ) 0 . 52 3 0 . 48 + ( 4 4 ) 0 . 52 4 . = 0 . 3431. (b) If the coin were fair, P ( X 3 ) = ( 4 3 ) 0 . 5 3 0 . 5 + ( 4 4 ) 0 . 5 4 = 0 . 3125. 5.9. (a) Separate ﬂips are independent (coins have no “memory,” so they do not try to compensate for a lack of tails). (b) Separate ﬂips are independent (coins have no “memory,” so they do not get on a “streak” of heads). (c) ˆ p can vary from one set of observed data to another; it is not a parameter. 169

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170 Chapter 5 Sampling Distributions 5.10. (a) X is a count; ˆ p is a proportion. (b) The given formula is the standard deviation for a binomial proportion . The variance for a binomial count is np ( 1 p ) . (c) The rule of thumb in the text is that np and n ( 1 p ) should both be at least 10. If p is close to 0 (or close to 1), n = 1000 might not satisfy this rule of thumb. (See also the solution to Exercise 5.18.) 5.11. (a) A B ( 200 , p ) distribution seems reasonable for this setting (even though we do not know what p is). (b) This setting is not binomial; there is no fixed value of n . (c) A B ( 500 , 1 / 12 ) distribution seems appropriate for this setting. 5.12. (a) This is not binomial; X is not a count of successes. (b) A B ( 20 , p ) distribution seems reasonable, where p (unknown) is the probability of a defective pair. (c) This should be (at least approximately) a B ( n , p ) distribution, where n is the number of students in our sample, and p is the probability that a randomly-chosen student eats at least five servings of fruits and vegetables.
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IPS6e.ISM.Ch05 - Chapter 5 Solutions 5.1(a n = 1500(the...

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