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Mean and Standard Deviation...please help!?
A lawn care business is concerned about the number of complaint calls it has been receiving and want to address the issue.
Number of complaint calls...Frequency
0 to 1...3
2 to 3...9
4 to 5...15
6 to 7...10
8 to 9...3
Find the mean and standard deviation.
1 Answer
- Julius NLv 78 years agoFavorite Answer
Data: 0.5{3}, 2.5{9}, 4.5{15}, 6.5{10}, 8.5{3}
n= 40; Σx= 182; Σx²= 1000; Σx³= 6096.5; Σx^4= 40013.5
Mean(μ)= 4.55; median= 4.5; Mode= 4.5
σ= 2.073; Variance=σ²= 4.2975; CV% = 45.6%
σn-1 = 2.0995; (σn-1)²= 4.4077; CV% = 46.1%
