CH%202%20Solutions[1]

# CH%202%20Solutions[1] - Chapter 2 Solutions 2.1. The mean...

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Chapter 2 Solutions 2.1. The mean is x = 30 , 841 pounds. Only 6 of the 20 pieces of wood had breaking strengths below the mean. The distribution is skewed to the left, which makes the mean smaller than the “middle” of the set of numbers (the median). 2.2. The mean is 31.25 minutes, while the median is 22.5 minutes. This is what we expect for a right-skewed distribution like this one. 2.3. The median is \$216,200, and the mean is \$265,000. The distribution of housing prices will be right-skewed, so the mean will be higher. 2.4. With all seasons included, x = 37 and M = 37 home runs. With the outlier removed, x = 35 (down 2) and M = 35 . 5 home runs (down 1.5). Means are more sensitive to outliers, while medians are resistant to them. 2.5. (a) The Fve-number summary (all quantities in units of pounds) is Min = 23 , 040, Q 1 = 30 , 315, M = 31 , 975, Q 3 = 32 , 710, Max = 33 , 650. (b) Note the distances between the numbers in the Fve-number summary: In order, the gaps are 7275, 1660, 735, and 940 pounds. That the Frst two gaps are larger gives some indication of the left skew. 2.6. (a) The stock fund varied between about 1 . 7% and 1.9%. (b) The median return for both funds was about 0.1%. (c) The stock fund is much more variable—it has higher positive returns, but also lower negative returns. 2.7. No (barely): The IQR is Q 3 Q 1 = 30 10 = 20 minutes, so we would consider any numbers greater than 60 minutes to be outliers. 2.8. (a) The Fve-number summary is Min = 5 . 7%, Q 1 = 11 . 7%, M = 12 . 75%, Q 3 = 13 . 5%, Max = 17 . 6%. (b) Yes: The IQR is Q 3 Q 1 = 13 . 5% 11 . 7% = 1 . 8%, so we would consider to be outliers any numbers below 11 . 7% 2 . 7% = 9% or above 13 . 5% + 2 . 7% = 16 . 2%. Along with ±lorida and Alaska, Utah is an outlier (8.5% older residents). 2.9. (a) x = 32 . 4 6 = 5 . 4mgof phosphate per deciliter of blood. (b) The details of the computation are shown on the right. The standard deviation is s = q 2 . 06 5 . = 0 . 6419 mg / dl. x i x i x ( x i x ) 2 5.6 0.2 0.04 5.2 0.2 0.04 4.6 0.8 0.64 4.9 0.5 0.25 5.7 0.3 0.09 6.4 1 1 32.4 0 2.06 72

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Solutions 73 2.10. The means and standard deviations are basically the same: For set A, x A . = 7 . 501 and s A . = 2 . 032, while for set B, x B . = 7 . 501 and s B . = 2 . 031. Set A is left-skewed, while set B has a high outlier. Set A 3 1 4 7 5 6 1 7 2 8 1177 9 112 Set B 5 257 6 58 7 079 8 48 9 10 11 12 5 2.11. (a) Not appropriate: The distribution of percents of college graduates has a high outlier. (If we can justify removing the outlier, the distribution is reasonably symmetric, so x and s would be ±ne for the remaining data.) (b) x and s are ±ne: The Iowa Test score distribution is quite symmetric and has no outliers. (c) Not appropriate: The wood breaking-strength distribution is strongly skewed. 2.12. State: How does logging affect tree count? Formulate: We need to compare the distributions, including appropriate measures of center and spread. Solve: Stemplots are shown below. Based on these, x and s are reasonable choices; the means and standard deviations (in units of trees) are given in the table (below, right).
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## This note was uploaded on 12/30/2009 for the course PSYC PSYC 60 taught by Professor ? during the Fall '09 term at UCSD.

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CH%202%20Solutions[1] - Chapter 2 Solutions 2.1. The mean...

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