Question

A sample set is as the following with a missing value x. It is known that-18 x1 101212 12 141414 1414 16 16 18 1820 2020 22 3035 What is the missing data value of the data set? What is the median? What is the variance? 9.What is the variance, if this is a population? 10. Does it skew? (Right, Left, or Not) A population is listed as the followin 10 15 11. 12. 13. 14. 15. 16. Find μ =? Find Σ(x-A)-? Find Σ(x-μ)2 ? Find ΣΧ_? Find ΣΧ2- Find σ-? A population is listed as the followin 15 47 48 40 49 43 Find the 2 x Find the(292. 45 17. 18. 19. Find thex2) 20.What is the standard deviation of the population?

Answer 1

$6.~(x_1+10+3*12+5*14+2*16+2*18+3*20+22+30+35)=20*18\\ or,~x_1=20*18-(10+3*12+5*14+2*16+2*18+3*20+22+30+35)\\ =29\\ 7.~Arrange~the~data~in~increasing~order:\\ 10,12,12,12,14,14,14,14,14,16,16,18,18,20,20,20,22,29,30,35\\ Median=(10^{th}~observation+11^{th}~observation)/2=(16+16)/2=16\\ 8.~\sum x^2=7322\\ Variance=\frac{1}{20-1}(\sum x^2-20\bar{x}^2)=(7322-20*18^2)/(20-1)=44.3158\\ 9.~Variance=\frac{1}{20}(\sum x^2-20\bar{x}^2)=(7322-20*18^2)/20=42.1\\ 10.~\mu_3=\frac{1}{20}\sum(x_i-\bar{x})^3=328.2\\ \mu_2=42.1\\ \gamma_1=\frac{\mu_3}{\mu^{\frac{3}{2}}_2}=1.2015>0\\ Hence~skewed~at~right.\\ 11.~\mu=(1+3+6+10+15+19)/6=9\\ 12.~\sum(x-\mu)=0\\ 15.~\sum x^2=1^2+3^2+6^2+10^2+15^2+19^2=732\\ 13.~\sum(x-\mu)^2=\sum x^2-6\mu^2=732-6*9^2=246\\ 14.~\sum x=1+3+6+10+15+19=54\\ 16.~\sigma=\sqrt{\frac{1}{6}\sum(x-\mu)^2}=6.4031$

$17.~\sum x^2=(43^2+15^2+7^2+47^2+3^2+48^2+40^2+18^2+45^2+49^2)\\ =12995\\ 18.~(\sum x)^2=99225\\ 19.~\sum x^2-\frac{(\sum x)^2}{n}=12995-99225/10=3072.5\\ 20.~SD=\sqrt{3072.5}=55.4301$