Assume the average selling price for houses in a certain county is $321000 with a standard deviation of $34000.
a) Determine the coefficient of variation.
b) Caculate the z-score for a house that sells for $358000.
c) Using the Empirical Rule, determine the range of prices that includes 68% of the homes around the mean.
d) Using Chebychev's Theorem, determine the range of prices that includes at least 93% of the homes around the mean.
Answer:
Given,
mean = = $321000
Standard deviation = = $34000
a)
To determine the coefficient of variation
Here coefficient of variation = /
substitute values
= 34000/321000
= 0.1059
b)
To calculate the z score for a house that sells for $358000
Z - score = (X - )/
= (358000 - 321000) / 34000
= 37000/34000
= 1.0882
c)
To determine the range of prices that includes 68% of home around mean
Now by utilizing empirical formula,
68% is within the 1 one standard deviation of mean
68% is within 321000 +/- 34000
= (321000 - 34000 . 321000 + 34000)
= (287000 , 355000)
d)
To determine the range of prices that includes 93% of homes around mean
proportion of data within the k standard deviation of mean is given as follows
= (1 - 1/k^2)
So (1 - 1/k^2) = 0.93
1/k^2 = 1 - 0.93
1/k^2 = 0.07
k^2 = 1/.07
k^2 = 14.2857
k = 3.7796
So at least 93% is in interval = mean +/- k*standard deviation
substitute values
= 321000 +/- 3.7796*34000
= 321000 +/- 128506.4
= (321000 - 128506.4 , 321000 + 128506.4)
= (192494 , 449507)
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