Statistics Homework Solutions 103

Statistics Homework Solutions 103 - 103 SECTION 4.11 ¡¥...

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Unformatted text preview: 103 SECTION 4.11 ¡¥ X100 be the heights of the 100 men. 0 25. 2 00. 0 9772. ¨ ¡ ¡ ¡ ¨ 0 0228 ¨ £ ©¥ ¨ ¡ ¨ ¡ ¤ ¡ ¡ 0 9772. ¨ ¨ ¡ ¨ ¡ 0 030187. 1 99. ¨ ¡ ¡ ¨ £ ¥ ¨ 1 99 is 0 0233. ¨ ¨ ¡ ¨ ¡ ¡ ¤ 18 80 ¤£ The area to the left of z 1 86 0 030187 0 27 ¡ ¨¨¨  ¤ The z-score of 1.8 is 1 8 1 86 and σX ¨ ¡ ¡¥¨ X80 be the breaking strengths of the 80 fabric pieces. Then X is approximately normally distributed with mean µX 0 0233. ¨ ¡¥¨ PX 100 ¨ 5. (a) Let X1 ¡ 69 5 ¡ PX 70 0 25 2 00 is 1 ¨ The area to the right of z ¤£ ¨¨¨  The z-score of 69.5 is 69 5 25 ¡ 70 and σX Then X is approximately normally distributed with mean µX ¨ ¤ ¡ Let X1 0 0170. ¨ 3. ¡ 100 ¡ PX 2 12 is 0.0170. ¨ The area to the left of z ¤ ¡ (b) Let x80 denote the 80th percentile 1 86 0 030187. ¨ £ ©¥ ¨ ¡ ¤ ¡ ¨ 1 8854 mm. ¨ ¤£ 18 0 01, 1.8 is the 1st percentile of the distribution of X . ¨ n . Solving for n yields n ¥ ¤£ 2 33 0 27 ¡ ¨ ¤ ¨ 1 86 ¨ ¤ ¡ ¨ ¡ ¡ Therefore 1 8 2 33. ¡¥¨ ¡ The z-score of the 1st percentile is approximately z ¡ ¨ Since P X ¨ 110. From the results of Example 4.70, the probability that a randomly chosen wire has no flaws is 0.48. Let X be the number of wires in a sample of 225 that have no flaws. ¨ ¤ ¡ ¡¥ ¨ ¤ ¡ ¥ ¨  ¤ ¥ 56 16 ¨ £ ¤ ©¥ ¨ ¡ ¤ ¤ 0 20 is 0.5793. ¡ 108 0 20. ¨ ¨ ¡ 0 5793. ¡ The area to the left of z ¨ ¡¥ 110 225 0 48 0 52 110 , use the continuity correction and find the z-score of 109.5. The z-score of 109.5 is 109 5 ¡ PX 108, and σ2 X ¤¥ To find P X 225 0 48 ¨ Bin 225 0 48 , so µX ¡¥ Then X 56 16. ¨ ¤ 7. ¡ (c) Let n be the necessary sample size. Then X is approximately normally distributed with mean µX σX 0 27 n. 1 86 and ¨ ¡ x80 x80 ¡ Therefore x80 satisfies the equation 0 84 0 84. ¨ The z-score of the 80th percentile is approximately z ...
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