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The accompanying frequency distribution represents the square footage of a random sample of 500 houses that are owner occupied year round. Approximate the mean and standard deviation square footage. statcrunch
Square footage Frequency 0- 499 500 999 13 1,000 1.499 33 1,500 1.999 115 2,000- 2.499 125 2,500 2.999 81 3,000- 3.499 3,500- 3.999 45 4,000 4.499 22 4,500 4.999 10

Answer :

Answer:

[tex]\bar X = \frac{\sum_{i=1}^n x_i f_i}{n} = \frac{1220750}{500}=2441.5[/tex]

[tex] s= \sqrt{\frac{N \sum x^2 f -[\sum xf]^2}{N(N-1}}= \sqrt{\frac{500*3408029125 -[1220750]^2}{50*49}}=9341.2405[/tex]

Step-by-step explanation:

In order to find the mean and standard deviation we can create the following table:

Limits            Frequency(f)      x(midpoint)      x*f           x^2 *f

__________________________________________________

0-499                   9                   249.5         2245.5    560252.3

500-999              13                  749.5         9743.5     7302753

1000-1499           33                 1249.5       41233.5    51521258.25

1500-1999           115                1749.5       201192.5   351986278.8

2000-2499         125               2249.5      281187.5   632531281.3

2500-2999          81                2749.5      222709.5  612339770.3

3000-3499          47                3249.5      152726.5   496284761.8

3500-3999          45                3749.5      168727.5   632643761.3

4000-4499          22                4249.5      93489       397281505.5

4500-4999          10                 4749.5       47495      225577502.5

_____________________________________________________

Total                    500                               1220750     3408029125

We can calculate the mean with the following formula:

[tex]\bar X = \frac{\sum_{i=1}^n x_i f_i}{n} = \frac{1220750}{500}=2441.5[/tex]

And the standard deviation would be given by:

[tex] s= \sqrt{\frac{N \sum x^2 f -[\sum xf]^2}{N(N-1}}= \sqrt{\frac{500*3408029125 -[1220750]^2}{50*49}}=9341.2405[/tex]

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