A population comprises numbers 1,3,5,7,9 and 11. Consider all population of size two when sampling with replacement.
Find the
1. The mean of population
2. The standard deviation of the population.
3. The mean of the sampling distribution of means.
4. The standard deviation of the sampling distribution of means.
N=6
n=2
Number of samples =NnN^nNn= 62= 36
μ=1+3+5+7+9+116=6\mu = \frac{1+3+5+7+9 +11}{6} = 6μ=61+3+5+7+9+11=6
σ=∑(x−μ)2Nσ=(1−6)2+(3−6)2+(5−6)2+(7−6)2+(9−6)2+(11−6)26=3.415\sigma = \sqrt{\frac{\sum (x - \mu)^2}{N}} \\ \sigma = \sqrt{\frac{(1-6)^2+(3-6)^2+(5-6)^2 +(7-6)^2+(9-6)^2+(11-6)^2}{6}} = 3.415σ=N∑(x−μ)2σ=6(1−6)2+(3−6)2+(5−6)2+(7−6)2+(9−6)2+(11−6)2=3.415
xˉ=1+2+...+10+1136=6\bar{x} = \frac{1+2+...+10+11}{36} = 6xˉ=361+2+...+10+11=6
s=(1−6)2+(2−6)2+...+(10−6)2+(11−6)236=2.4495s = \sqrt{\frac{(1-6)^2+(2-6)^2+...+(10-6)^2 +(11-6)^2}{36}} = 2.4495s=36(1−6)2+(2−6)2+...+(10−6)2+(11−6)2=2.4495
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