A sample of 5 hotels reported the price for a night's stay:$189 $219 $259 $329 $129What is the range of the prices of a night’s stay? How to find the standard deviation of the nightly prices
Accepted Solution
A:
Answer:a) Range = 200b) Standard Deviation = 75.03Step-by-step explanation:Prices: $189 $219 $259 $329 $129a) Range Range = Maximum value - Minimum valueMaximum value = 329Minimum value = 129So, Range = 329 - 129Range = 200b) Standard DeviationThe formula used for finding standard deviation is:[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_{i}-\mu)^2}[/tex]N is No of termsμ is meanMean μ = (189+219+259+329+129)/5 = 225x x-μ (x-μ)^2189 189-225= -36 1296219 219 - 225= -6 36 259 259-225= 34 1156 329 329-225=104 10,816129 129-225=-96 9216Now, find [tex]\sum_{i=1}^N(x_{i}-\mu)^2[/tex][tex]\sum_{i=1}^N(x_{i}-\mu)^2=1296+36+1156+10816+9216[/tex][tex]\sum_{i=1}^N(x_{i}-\mu)^2=22520[/tex]Now finding standard deviation [tex]\sigma[/tex][tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_{i}-\mu)^2}[/tex][tex]\sigma=\sqrt{\frac{1}{5-1}(22520)}\\\sigma=\sqrt{5630}\\\sigma=75.03[/tex]