Statistics Homework Solutions 89

Statistics Homework Solutions 89 - 89 SECTION 4.6 ¨ ¡ ¡...

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Unformatted text preview: 89 SECTION 4.6 ¨ ¡ ¡ ¨ ¥ 0 25 ¡ £ ¤ ©¥ ¨ ¡ ¨  ¡¥ ¤ ¤ 1 00. 0 1587. ¨ ¡ ¨ ¨ ¡ ¡ 0 1587. ¨ ¥  ¤ 0 01. The z-score of 0.5 is 0 5 0 σ. £ ¥ ¨ ¤ ¨ ¡ 2 33. Solving for σ yields σ 0 2146, and σ2 ¨ ¨ ¤ ¤ ¡¥ ¡ ¤ ¡ ¡¥¨ ¨ ¡ ¤ ¡ ¤ £ ©¥ ¡ ¡¥ ¡ ¡¥¨ 0σ ¡ ¡ ¡¥ 0 01, the z-score of 0.5 is 2.33. 0 04605. ¨ ¡ 05 N 0 σ2 . ¡ ¡ ¥¨ 0 8413 0 ¨ ¨ 1 00 is 1 E , so X 05 ¡ ¤ ¤ Therefore 0 5 ¨ Since P X ¡ 0, then X PX N 0 0 25 . 0 5 . The z-score of 0.5 is 0 5 Therefore P error P error 0 0485. E , so X The area to the right of z (b) If m 0 0485. ¨ ¤ PX ¡ P error £ ¥ 0, then X ¨ 23. (a) If m 2000 0 9515 1 66. ¡ 1 66 is 1 The area to the right of z Therefore P S 1700 180 278 ¨ The z-score of 2000 is 2000 ¨ ¤ 25. (a) The sample mean is 114.8 J and the sample standard deviation is 5.006 J. (b) The z-score of 100 is 100 114 8 5 006 2 96. The area to the left of z 2 96 is 0.0015. Therefore only 0.15% of bolts would have breaking torques less than 100 J, so the shipment would be accepted. ¡ ¡ ¨ ¡ ¡ ¨ £ ©¥ ¨ ¡ ¤ ¨ £ ©¥ ¨ ¡ ¤ ¨ ¡ ¡ ¨ ¡ (d) The bolts in part (c) are stronger. In fact, the weakest bolt in part (c) is stronger than the weakest bolt in part (a), the second-weakest bolt in part (c) is stronger than the second-weakest bolt in part (a), and so on. (e) The method is certainly not valid for the bolts in part (c). This sample contains an outlier (140), so the normal distribution should not be used. Section 4.6 1. Let Y be the lifetime of the component. ¡ ¨ (c) The sample mean is 117.08 J; the sample standard deviation is 8.295 J. The z-score of 100 is 100 117 08 8 295 2 06. The area to the left of z 2 06 is 0.0197. Therefore about 2% of the bolts would have breaking torques less than 100 J, so the shipment would not be accepted. ...
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