Question

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.

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 1

Answer:

$\bar X = \frac{\sum x_i f_i}{n} = \frac{1220750}{500}=2441.5$

$s^2= \frac{3408029125 -\frac{(1220750)^2}{500}}{500-1} =856849.7$

$s= \sqrt{856849.7}=925.662$

Step-by-step explanation:

For this case we can create the following table

Interval      Frequency (f)    Midpoint(xi)       xi *f      xi^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          201193.5  361986278.8

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

$\sum f_i = 500 , \sum x_i f_i = 1220750, \sum x^2_i f_i = 3408029125$

For this case we can calculate the sample mean with this formula:

$\bar X = \frac{\sum x_i f_i}{n} = \frac{1220750}{500}=2441.5$

And for the sample variance we can use the following formula:

$s^2= \frac{\sum x^2_i f_i - \frac{(\sum x_i f_i)^2}{n}}{n-1}$

And if we replace we got:

$s^2= \frac{3408029125 -\frac{(1220750)^2}{500}}{500-1} =856849.7$

And the deviation is just the square root of the sample variance and for this case is:

$s= \sqrt{856849.7}=925.662$

Answer 2

The mean of a distribution is its average, while the standard deviation is the summary of how far each data in the dataset, is to the mean.

The mean and the standard deviation are 2441.5 and 924.7 respectively.

First, we calculate the class midpoint (x) from the given intervals.

The class midpoint is the average of the intervals. So, we have:

$x_1 = \frac{0+499}{2} = 249.5$

$x_2 = \frac{500+999}{2} = 749.5$

....

$x_{10} = \frac{4500+4999}{2} = 4749.5$

So, the table becomes:

$\begin{array}{cc}x&f&249.5&9&749.5&13&1249.5&33&1749.5&115&2249.5&125&2749.5&81&3249.5&47&3749.5&45&4249.5&22&4749.5&10\end{array}$

The mean of the distribution is:

$\bar x = \frac{\sum fx}{\sum f}$

This gives:

$\bar x = \frac{9 \times 249.5 + 12 \times 749.5 + 33 \times 1249.5 + 115 \times 1749.5 + 125 \times 2249.5 + .... + 10 \times 4749.5}{9+13+33+115+125+81+47+45+22+10}$

$\bar x = \frac{1220750}{500}$

$\bar x = 2441.5$

The standard deviation is calculated as follows:

$\sigma = \sqrt{\frac{\sum f(x - \bar x)^2}{\sum f}}$

So, we have:

$\sigma = \sqrt{\frac{9 \times (249.5 -2441.5)^2 + 13 \times (749.5 -2441.5)^2+ 33 \times (1249.5 -2441.5)^2+...........+ 10 \times (4749.5-2441.5)^2}{9+13+33+115+125+81+47+45+22+10}$

$\sigma = \sqrt{\frac{427568000}{500}$

$\sigma = \sqrt{855136$

$\sigma = 924.7$

Hence, the mean and the standard deviation are 2441.5 and 924.7 respectively.

Read more about mean and standard deviation at:

https://brainly.com/question/10729938