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標題:
英文統計學mean standard deviation
發問:
可以幫下我嗎?2.A factory produces steel rods that are used to reinforce concrete. The production process is known to produce rods that are normally distributed with a mean length of 300 cm and a standard deviation of 8 cm.(a) One rod is randomly selected for quality checking. i.What is the probability that... 顯示更多 可以幫下我嗎? 2.A factory produces steel rods that are used to reinforce concrete. The production process is known to produce rods that are normally distributed with a mean length of 300 cm and a standard deviation of 8 cm. (a) One rod is randomly selected for quality checking. i.What is the probability that a randomly selected rod has a length longer than 315 cm? ii.If 22% of the rods are more than a specific length L, find L. (a)14 rods are randomly selected for quality checking. i.What are the mean and standard deviation of the sample mean length? ii.What is the probability that the sample mean length will be between 297 cm and 305 cm? iii.If 22% of the sample means are more than a specific length L, find L. 請列式=="
最佳解答:
(a) i P( X > 315 ) = P( Z > (315-300)/8 ) = P( Z > 1.875 ) 查標準常態分配機率表得 P( Z > 1.87 ) = 0.0307 , P( Z > 1.88 ) = 0.0301 故 P( Z > 1.875 ) = (0.0307+0.0301) / 2 = 0.0304 Ans: 0.0304 ii 0.22 = P( X > L ) = P( Z > (L-300)/8 ) 查表得P( Z > 0.77 ) = 0.22 故 0.77 = (L-300)/8 L = 300+8*0.77 = 306.16 (cm) Ans: L = 306.16 cm (a) 14 rods ...... 應該改成 (b)14 rods ...... (b) i μ of X-bar = μ = 300 σof X-bar = σ / √n = 8/√14≒ 2.138 Ans: The mean of the sample mean length = 300 cm The standard deviation of the sample mean length = 2.138 cm ii P( 297 L ) = P( Z > (L-300)/2.138 ) 查表得P( Z > 0.77 ) = 0.22 故 0.77 = (L-300)/2.138 L = 300+2.138*0.77 ≒301.65 (cm) Ans: L = 301.65 cm 註解 : 題目(b)沒說群體數大小,但鋼棒是用來強化混凝土的結構, 顯然使用數量會超過樣本數的10倍以上,可視為無限群體. 故計算x-bar的標準誤時,不用修正係數(或是說修正係數為1)
其他解答:
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