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Unformatted text preview: Stat 350 Solution 3 6. (a) The reported blood pressure values would have been: 120 125 140 130 115 120 110 130 135. When ordered these values are: 110 115 120 120 125 130 130 130 135 140. The median is 125. (b) 127.6 would have been reported to be 130 instead of 125. Since this value is the middle value, the median would change from 125 to 130. This example illustrates that the median is sensitive to rounding in the data. 9. (a) = 2 ) 5 (. dx x x = 2 3 3 5 . x = 4/3. The mean does not equal 1 because the density curve is not symmetric around x = 1. (b) Half the area under the density curve to the left (or right) of the median ~ , so, .50 = ~ 5 . dx x = ~ 2 2 5 . x = ( ~ 2 )/4, or, ~ 2 = 4(.5) = 2 and ~ = 1.414. < ~ because the density curve is negatively skewed. (c) 2 1 = 3 4 2 1 = 6 5 and 6 11 . The area under the curve between these two values is 6 / 11 6 / 5 5 . dx x = 6 / 11 6 / 5 2 4 x = .667. Similarly, the proportion of the times that are within onehalf hour of ~ is: 914 . 1 914 . 5 . dx x = 914 . 1 914 . 2 4 x = .707. 11. = 2 1 2 1 1 2 dx x x = dx x dx x 2 2 2 1 2 1 = 2 1 2 1 2 ) ln( 2 x x = (2 21 2 )  2(ln(2)ln(1)) = 3  2(.693147) = 1.61371, or about 1.614 . For the median, ~ , half of the area under the density must lie to the left of ~ , so .5 = dx x 1 2 1 1 2 = 1 1 2 2 x x = 2( ~1) + 2(1/ ~ 1). Solving for ~ results in a quadratic equation 2 ~ 2 4.5 ~ + 2 = 0, whose roots are .61 and 1.64; only the root ~ = 1.64 is feasible (since it lies in the interval from 1 to 2). The proportion of x values that lie within the mean and median is dx x 1 1 2 64...
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 Spring '08
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 Statistics

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