Question #266761

Present a complete process in computing the mean of the sampling distribution of the sample


means given the General Mathematics 20- item quiz result of 6 students; 12, 16, 18, 20, 10, 15.


Consider a sample size of 5.

1
Expert's answer
2021-11-17T02:40:22-0500

Solution:

Given result of 6 students; 12, 16, 18, 20, 10, 15.

Here, a sample size of 5, so n=5

We consider no repetition.

Samples are: (12, 16, 18, 20, 10), (16, 18, 20, 10, 15), (12, 18, 20, 10, 15), (12, 16, 20, 10, 15), (12, 16, 18,10, 15), (12, 16, 18, 20,15).

Their means are:

Mean of (12, 16, 18, 20, 10)=(12+16+18+20+10)/5=15.2=(12+16+18+20+10)/5=15.2

Mean of (16, 18, 20, 10, 15)=(16+18+20+10+15)/5=15.8=(16+18+20+10+15)/5=15.8

Mean of (12, 18, 20, 10, 15)=(12+15+18+20+10)/5=15=(12+15+18+20+10)/5=15

Mean of (12, 16, 20, 10, 15) =(12+16+15+20+10)/5=14.6=(12+16+15+20+10)/5=14.6

Mean of (12, 16, 18,10, 15) =(12+16+18+15+10)/5=14.2=(12+16+18+15+10)/5=14.2

Mean of (12, 16, 18, 20,15)=(12+16+18+20+15)/5=16.2=(12+16+18+20+15)/5=16.2

Now, mean of sampling distribution =(15.2+15.8+15+14.6+14.2+16.2)/6=1516=(15.2+15.8+15+14.6+14.2+16.2)/6=15\dfrac16


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