5 2 34028 c 2 2 6 100 using the results found in

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Unformatted text preview: n −1 = b2 sx 2 y = −32 + 5 / 9 x = −32 + 5 / 9(835.00) = 431.89 °C s y = b 2 s x = (5 / 9) 2 (10.5) 2 = 34.028 °C 2 2 6-100. Using the results found in Exercise 6-98 with a = − x and b = 1/s, the mean and standard deviation s of the zi are z = 0 and sZ = 1. 6-101. Yes, in this case, since no upper bound on the last electronic component is available, use a measure of central location that is not dependent on this value. That measure is the median. Sample Median = x ( 4 ) + x (5) 2 = 63 + 75 = 69 hours 2 6-60 n +1 6-102 a) x n +1 = x n +1 x n +1 b) n ∑ xi ∑x i =1 i =1 i + x n +1 = n +1 n +1 nx n + x n +1 = n +1 x n = x n + n +1 n +1 n +1 2 ns n +1 ⎛n ⎞ ⎜ ∑ xi + x n +1 ⎟ n ⎠ 2 = ∑ xi2 + x n +1 − ⎝ i =1 n +1 i =1 2 2 n ⎛n ⎞ ⎜ ∑ xi ⎟ 2 x n +1 ∑ xi n x2 i =1 ⎠− 2 i =1 − n +1 = ∑ xi2 + x n +1 − ⎝ n +1 n +1 n +1 i =1 ⎛n ⎞ ⎜ ∑ xi ⎟ n n n2 2 2x n +1 x n − ⎝ i =1 ⎠ x n +1 − = ∑ xi + n +1 n +1 n +1 i =1 2 ⎡n (∑ xi )...
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This document was uploaded on 02/09/2014.

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