SUPPOSE THE PRICES OF A CERTAIN MODEL OF NEW HOMES ARE NORMALLY DISTRIBUTED WITH A MEAN OF $150,000. uSE THE 68-95-99.7 RULE TO FIND THE PERCENTAGE OF BUYERS WHO PAID BETWEEN $150,000 AND $152,700 IF THE STANDARD DEVIATION IS $900.00
In the case of a normal distribution with a mean m and a standard deviation d, you have the following property :
P((m-d<X<m+d) around 68,27% (1)
P((m-2d<X<m+2d) around 95,45% (2)
P((m-3d<X<m+3d) around 99,73% (3)
In this case m=150,000 and d=900 or 2,700=3d, so you are asked to provide the percentage of buyers paying between m and m+3d.
The probability density function is symetrical around m so P((m<X<m+3d) (what's aked you to provide) = P((m-3d<X<m+3d)/2=49,865%
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