A.
Minimum | = MIN(E14:E19) | 8 |
Maximum | = MAX(E14:E19) | 18 |
First quartile, Q1 | = QUARTILE.EXC(E14:E19,1) | 9.5 |
Third quartile, Q3 | = QUARTILE.EXC(E14:E19,3) | 14.25 |
IQR | = Q3-Q1 | 4.75 |
B. Lower limit = Q1 -1.5IQR = 9.5 - 1.5*4.75 = 2.375
Upper limit = Q3 + 1.5IQR = 14.25 + 1.5*4.75 = 21.375
As all the value are above lower limit and below upper limit, we can say that there are no outliers.
Styles 1. The expected life expectancy for a sample of 6 dog breeds was collected. Breed...
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TATS lessen 2 Saved to this PC References MailingsReview View Help Tell me what you want to do AaBbCcD AaBbCcD AaBbC AaBbCcl A Normal No Spac... Heading 1 Heading 2 Styles 1. The expected life expectancy for a sample of 6 dog breeds was collected. Breed Collie Cockapoo Dachshund Life Expectancy 18 12 English Bulldog 10 Pointer Malamute 13 A. By hand, calculate the mean...