A population consists of four numbers 3, 7, 11, 15. consider all possible sample size 2 which can be drawn without replacement. Find mean and SD in the population.
Number=N!n!(N−n)!=4!2!(4−2)!=3×42=6Number = \frac{N!}{n!(N-n)!} \\ = \frac{4!}{2!(4-2)!} \\ = \frac{3 \times 4}{2} \\ = 6Number=n!(N−n)!N!=2!(4−2)!4!=23×4=6
Samples:
3,7
3,11
3,15
7,11
7,15
11,15
Mean=3+7+11+154=9SD=14−1((3−9)2+(7−9)2+(11−9)2+(15−9)2)=13(36+4+4+36)=5.164Mean = \frac{3+7+11+15}{4} = 9 \\ SD = \sqrt{\frac{1}{4-1}( (3-9)^2+(7-9)^2 +(11-9)^2 + (15-9)^2 )} \\ = \sqrt{ \frac{1}{3}(36+4+4+36) } \\ = 5.164Mean=43+7+11+15=9SD=4−11((3−9)2+(7−9)2+(11−9)2+(15−9)2)=31(36+4+4+36)=5.164
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment