Use following sample home prices ($000s) to answer the questions:
175.0 48.8 995.9 298.0 218.9 209.0
628.3 111.0 212.3 2325.0 958.0 212.6
a) What is the sample median home price?
b) Do the data contain any outliers?
c) Determine variance and standard deviation.
d) Develop a Box Plot for the home prices. ]
e) Determine 8th decile.
f) Determine the coefficient of skewness using the software method.
g) In the given case, which is the better measure of Central Tendency? Why?
h) At least what percent of home prices will lie between two standard deviations of mean?
Urgent need please
Firstly we will write the given Observation in ascending order-
"48.8, 111, 175,209,212.3,212.6,218.9,298,628.3,958,995.9,2325.0"
Total number of terms n=12
(i) Medians are ="(\\dfrac{n}{2})th" ,"(\\dfrac{n}{2}+1)th" terms
="(\\dfrac{12}{2})th,(\\dfrac{12}{2}+1)th" terms
="6^{th},7^{th}" terms
="212.6,218.9"
Average median ="\\dfrac{212.6+218.9}{2}=\\dfrac{431.5}{2}=215.75"
(ii) Yes There are total of three outliers i.e. "958,995,2325" respectively.
(iii) Mean ="\\dfrac{\\sum X}{n}=\\dfrac{6284.76}{12}=523.73"
Variance "\\sigma^2" ="\\dfrac{\\sum(\\bar{X}-X)^2}{12}=\\dfrac{4719445.3}{12}=393287"
Standard deviation "\\sigma=\\sqrt{\\sigma^2}=\\sqrt{393287}=627.12"
(iv) Box plot for the fiven set of data is given by-
(v) The formula for decile is given by-
"D_8=\\dfrac{k}{10}(N+1)" term
where, k= Decile term
N=Total no. of terms
"D_8=\\dfrac{8}{10}(12+1)=0.8\\times 13=10.4=10^{th}" term
Hence "D_8=958.0"
(vi) The formula for coefficient of skewness is given by-
skewness= "\\dfrac{\\sum(y_i-y)^3}{(n-1)\\times (SD)^3}"
= 1.5351
(vii) In this case Median is the better measure of central tendency Since There are any outliers
present in the Given set of data ,
In median Their value can be moderate.
(viii) Since As the study suggest, Generally There are 95% people who tell the truth,
There are approximate 3% who told lie about their home price.
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