close
標題:
NSS MI (Statistic Part) 急急急
發問:
最佳解答:
If the mark is 50, then the toal mark ∑ x_i =38*65=2470. Also ∑ (x_i - 65)^2 / 38 = (2√2)^2 => ∑ (x_i - 65)^2 = 304 ∑ (x_i)^2 - 130 ∑x_i + 38*4225 = 304 ∑ (x_i)^2 =160854 Now one recorded is changed to 60. So the new ∑ x_i is 2470-50+60 = 2480 and the mean is 2480/38 = 65.2632 New ∑ (x_i)^2 = 160854 - 50^2 + 60^2 = 161954 New Var(x) = E(x^2)-[E(x)]^2 = 161954 / 38 - [2480 / 38]^2 = 2.6676 So the new standard deviation = 1.633
其他解答:
NSS MI (Statistic Part) 急急急
發問:
此文章來自奇摩知識+如有不便請留言告知
Question:The mean and the standard deviation of the Liberal Studies test marks of 38 students are 65 and 2√2 respectively. After checking the marks, the teacher finds that a mark of 60 is wrongly recorded as 50. Find the correct mean and standard deviation of the test marks.I have already the mean but i... 顯示更多 Question: The mean and the standard deviation of the Liberal Studies test marks of 38 students are 65 and 2√2 respectively. After checking the marks, the teacher finds that a mark of 60 is wrongly recorded as 50. Find the correct mean and standard deviation of the test marks. I have already the mean but i don't know how to find standard deviation. The mean is 65.26(4 sig. fig.) Sources: Mathematics In Action Module 1 Calculus and Statistic (Volume 2) (Longman) Chapter 9 Review on Basic Statistics and Probability (Exercise 9A no. 19) P. 9.24最佳解答:
If the mark is 50, then the toal mark ∑ x_i =38*65=2470. Also ∑ (x_i - 65)^2 / 38 = (2√2)^2 => ∑ (x_i - 65)^2 = 304 ∑ (x_i)^2 - 130 ∑x_i + 38*4225 = 304 ∑ (x_i)^2 =160854 Now one recorded is changed to 60. So the new ∑ x_i is 2470-50+60 = 2480 and the mean is 2480/38 = 65.2632 New ∑ (x_i)^2 = 160854 - 50^2 + 60^2 = 161954 New Var(x) = E(x^2)-[E(x)]^2 = 161954 / 38 - [2480 / 38]^2 = 2.6676 So the new standard deviation = 1.633
其他解答:
文章標籤
全站熱搜
留言列表
