# Hw2 - STAT 350 Homework #2 SOLUTION Text Exercises Chapter...

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STAT 350 – Homework #2 SOLUTION Text Exercises Chapter 1, Problem #40 1.40 X ~ Normal( μ = 8.8, σ = 2.8) a. P(X > 10) = P(Z > (10-8.8)/2.8) = P(Z > 0.4286) = 1 - Φ (0.43) = 1 - 0.6664 = 0.3336 P(X 10) = P(X > 10) = 0.3336 b. P(X 5) = P(Z (5-8.8)/2.8) = P(Z < -1.357) = Φ (-1.36) = 0.0869 P(5 < X < 10) = P[(5-8.8)/2.8 < Z < (10-8.8)/2.8] = P(-1.36 < Z < -1.36) = Φ (0.43) - Φ (- 1.36) = 0.6664 - 0.0869 = 0.5795 c. P(X > 20) = P(Z > (20-8.8)/2.8) = P(Z > 4.00) = approximately zero (< 0.0001) d Φ (z = 2.33) = 0.9901, so c is 2.33 standard deviations above and below the mean, c = 2.33(2.8) = 6.524, c = 6.524, thus 98% of trees have DBH between 2.276 inches and 15.324 inches Chapter 2, Problems #8, 9, 10, 14, 24 2.8 =0 () 1 2 1 0.75 1 fx EX x x d x ⎡⎤ =− ⎣⎦ 1442 4 43 2.9 a. Mean: = 4/3 1.333 () ( ) 2 0 0.5 x xd x = 123 A graph of the PDF illustrates that the larger values in the range are more likely than the smaller values, pulling the mean above the midpoint. b. Median: 2 0 0.5 0.5 0.5 0.5 2 m m x =⇒ = , m = 21 . 4 1 4 The mean is smaller than the median. This occurs because the distribution is negatively skewed, pulling the mean value down. c. P(4/3 - 1/2 < X < 4/3 + 1/2) = 11/6 5/6 0.5 x dx = 0.6667 1.914 0.914 0.5 x dx = 0.7071 P(1.414 - 0.5 < X < 1.414 + 0.5) = 2.10 1 b a x d x ba ⎛⎞ = ⎜⎟ ⎝⎠ 14243 = 2 ab +

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2.14 a. = 2 () () 2 22 0 () 0.5 fx EX x xd x = 123 b. for a continuous RV: ( )() () () () () () () 1 x x a b xf xd x a f x b x f x d x a f xd x b x x a f xd x b x x ab μ +∞ +∞ −∞ −∞ +∞ +∞ +∞ +∞ −∞ −∞ −∞ −∞ == += + ⎡⎤ ⎣⎦ =+ = + ∫∫ 142 43 1 42 43 OR for a discrete RV: ( ) ( ) ( ) 1 x x xpx a px b x px a b x a px b x + = + ∑∑ 142 4 31 42 43 If the mean repair time is μ x = 0.5, the mean repair revenue is μ 25+40 x = 25 + 40 μ x = 45 2.24 a. () ( ) ( ) ( ) { ( ) { ( ) { ( ) { ( { 2 22222 0.4 0.1 0.1 0.1 0.3 Var 1.8 01 . 8 0 11 . 8 1 21 . 8 2 31 . 8 3 41 . 8 4 Xx p x ppppp =− +− ) = 2.96 2.96 SD X = =1.72 b. P(1.8 - 1.72 < X < 1.8 + 1.72) = P(0.8 < X < 3.52) = p(1) + p(2) + p(3) = 0.1 + 0.1 + 0.1 = 0.3 P(1.8 - 3 × 1.72 < X < 1.8 + 3 × 1.72) = P(-3.36 < X < 6.96) = p(0) + p(1) + p(2) + p(3) + p(4) = 1.0 Then answer the following Problems 1. For the following problems, use the dataset in the Excel spreadsheet from Lab #1 . a. Give the 5-number summary for this dataset. The lower quartile is the average of 0.92 and 1.61: 1.27 The median is the average of the 50 th and the 51 st smallest observations: 13.57 and 14.05: 13.81 The upper quartile is the average of 41.01 and 43.52: 42.27 Thus, the 5-number summary is: 0.00 – 1.27 – 13.81 – 42.27 – 96.36
b. Are there any outliers in this dataset? If so, give their values.

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## This note was uploaded on 03/09/2010 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue.

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Hw2 - STAT 350 Homework #2 SOLUTION Text Exercises Chapter...

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