Question #199376

construct all random samples consisting 2 observation from the given data you are asked to guess the average weight of 6th watermelons by taking a random sample without replacement from the population


1
Expert's answer
2021-05-28T10:15:16-0400

Watermelon A B C D E F

Weight (in pounds) 19 14 15 9 10 17


Random samples consisting 2 observation:

(19,14),(19,15),(19,10),(19,9),(19,17),(14,15),(14,9),(14,10),(19,14),(19,15),(19,10),(19,9),(19,17),(14,15),(14,9),(14,10),

(14.17),(15,9),(15,10),(15,17),(9,10),(9,17),(10,17)(14.17),(15,9),(15,10),(15,17),(9,10),(9,17),(10,17)


For sample size n=5:

(19,14,15,9,10),x=13.4(19, 14, 15, 9, 10),\overline{x}=13.4

(19,14,15,9,17),x=14.8(19, 14, 15, 9, 17),\overline{x}=14.8

(19,14,15,10,17),x=15(19, 14, 15, 10, 17),\overline{x}=15

(19,14,9,10,17),x=13.8(19, 14, 9, 10, 17),\overline{x}=13.8

(19,15,9,10,17),x=14(19, 15, 9, 10, 17),\overline{x}=14

(14,15,9,10,17),x=13(14, 15, 9, 10, 17),\overline{x}=13

 The average weight of 6th watermelons by taking a random sample without replacement:

μ=13.4+14.8+15+13.8+14+136=14 pounds\mu=\frac{13.4+14.8+15+13.8+14+13}{6}=14\ pounds


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