Use the real estate data that is give in question 1 and 2.And complete the Sampling Distributions – Real Estate questions 3 and 4.1. Review the data and for the purpose ofthis project please consider the 100 listing prices as a population. – Explain what your computed population mean and population standard deviation were.Answer -Population mean isan average of 100 listing prices, μ = (Σ * X)/ N= $67,118,514/100 = $671,185.14. The computedpopulation mean for the 100 listing prices was $671,185.14. Population Standard Deviation is a measure of how spreads out numbers are. Researchers use statistics to estimate parameters, refer to a numerical property of a population. The real estate population Standard Deviation was σ= $274,980.89.2. Divide the 100 listing prices into 10 samples of n=10 each. Each of your 10 samples will tend to be random if the first sample includes houses 1 through 10 on your spreadsheet, the second sample consists of houses 11 through 20, and so on.Compute the mean of each of the 10 samples and list them:Answer- n=10 (Sample 1: listings 1-10) Mean = $561,808.90 n=10 (Sample 2: listings 11-20) Mean = $602,175.00 n=10 (Sample 3: listings 21-30) Mean = $783,299.50 n=10 (Sample 4: listings 31-40) Mean = $738,140.00 n=10 (Sample 5: listings 41-50) Mean = $842,058.80 n=10 (Sample 6: listings 51-60) Mean = $817,098.80 n=10 (Sample 7: listings 61-70) Mean = $626,167.60 n=10 (Sample 8: listings 71-80) Mean = $671,445.00 n=10 (Sample 9: listings 81-90) Mean = $498,889.00 n=10 (Sample 10: listings 91-100) Mean = $570,768.80 3. Computethe mean of those 10 means.- Explainhow the mean of the means is equal, ornot, to the population mean of the 100 listing pricesfrom above.4. Computethe standard deviation of those 10 means and compare the standard deviation ofthe 10 means to the population standard deviation of all 100 listingprices.- Explainwhy it is significantly higher, or lower, than the population standarddeviation.For additional information use the attachment as reference.
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