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!! IM- || Done Topic 2 (2 of 4) loannis Kyriakou Introduction to Probability and Statistics Numerical Data Summaries 20/28 Example Calculate the sample mean and standard deviation of the following values: 534 671 581 620 401 340 980 845 550 690 • n= 10 =Σx-621.20 (SH675TH204 fotH:360+98043451550190)=621.2 = 621.202 = 385, 889.44 E x} = 4, 186, 784 i=1 = S O n-1 x? – nx2 = 190.87 - f 5.00. 21/28 Numerical Data Summaries loannis Kyriakou Introduction to Probability and Statistics Example ext page %D

!! IM- || Done Topic 2 (2 of 4) loannis Kyriakou Introduction to Probability and Statistics Numerical Data Summaries 20/28 Example Calculate the sample mean and standard deviation of the following values: 534 671 581 620 401 340 980 845 550 690 • n= 10 =Σx-621.20 (SH675TH204 fotH:360+98043451550190)=621.2 = 621.202 = 385, 889.44 E x} = 4, 186, 784 i=1 = S O n-1 x? – nx2 = 190.87 - f 5.00. 21/28 Numerical Data Summaries loannis Kyriakou Introduction to Probability and Statistics Example ext page %D

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.5: Comparing Sets Of Data
Problem 2CYU
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I was reading through my lecture notes and I'm not sure how they got 4,186,784. 

!! IM- || Done Topic 2 (2 of 4) loannis Kyriakou Introduction to Probability and Statistics Numerical Data Summaries 20/28 Example Calculate the sample mean and standard deviation of the following values: 534 671 581 620 401 340 980 845 550 690 • n= 10 =Σx-621.20 (SH675TH204 fotH:360+98043451550190)=621.2 = 621.202 = 385, 889.44 E x} = 4, 186, 784 i=1 = S O n-1 x? – nx2 = 190.87 - f 5.00. 21/28 Numerical Data Summaries loannis Kyriakou Introduction to Probability and Statistics Example ext page %D
Transcribed Image Text:!! IM- || Done Topic 2 (2 of 4) loannis Kyriakou Introduction to Probability and Statistics Numerical Data Summaries 20/28 Example Calculate the sample mean and standard deviation of the following values: 534 671 581 620 401 340 980 845 550 690 • n= 10 =Σx-621.20 (SH675TH204 fotH:360+98043451550190)=621.2 = 621.202 = 385, 889.44 E x} = 4, 186, 784 i=1 = S O n-1 x? – nx2 = 190.87 - f 5.00. 21/28 Numerical Data Summaries loannis Kyriakou Introduction to Probability and Statistics Example ext page %D
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