Airport 4

The airport in Košice collected the following data on aircraft delays within one week:
(Solve the following tasks without classification into intervals. Round the results to 2 decimal places)

Aircraft delay in min.: 541; 545; 575; 520; 572; 544; 524; 567; 569; 552; 585; 588; 535; 538; 549; 604; 540; 564; 593; 553

What value of the standard deviation of measurement can Košice Airport expect for flights during the examined week?

Final Answer:

σ =  23.3304 min

Step-by-step explanation:

s=541+545+575+520+572+544+524+567+569+552+585+588+535+538+549+604+540+564+593+553=11158 n=20  μ=s/n=11158/20=105579​=557.9 min  r=n−11​⋅ ((541−μ)2+(545−μ)2+(575−μ)2+(520−μ)2+(572−μ)2+(544−μ)2+(524−μ)2+(567−μ)2+(569−μ)2+(552−μ)2+(585−μ)2+(588−μ)2+(535−μ)2+(538−μ)2+(549−μ)2+(604−μ)2+(540−μ)2+(564−μ)2+(593−μ)2+(553−μ)2)=20−11​⋅ ((541−557.9)2+(545−557.9)2+(575−557.9)2+(520−557.9)2+(572−557.9)2+(544−557.9)2+(524−557.9)2+(567−557.9)2+(569−557.9)2+(552−557.9)2+(585−557.9)2+(588−557.9)2+(535−557.9)2+(538−557.9)2+(549−557.9)2+(604−557.9)2+(540−557.9)2+(564−557.9)2+(593−557.9)2+(553−557.9)2)=9551709​≐544.3053  σ=r​=544.3053​=23.3304 min

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