Question

The market research department of the National Real Estate Company conducted a survey among 500 prospective buyers in a suburb of a large metropolitan area to determine the maximum price a prospective buyer would be willing to pay for a house. From the data collected, the distribution that follows was obtained.

Maximum Price Considered, x P(X = x)
(in thousands of dollars)
480 10
500
490 20
500
500 75
500
510 85
500
520 70
500
550 95
500
580 95
500
600 45
500
650 5
500


Required:
Compute the mean, variance, and standard deviation of the maximum price x that these buyers were willing to pay for a house.

Answer 1

Answer:

Following are the responses to the given question:

Explanation:

Using formula:

$Mean=\Sigma x \times P(x)\\\\Variance=\Sigma (x-\mu)^2 \times P(x)\\\\Standard \ Deviation=\sqrt{Variance}$

$x\ \ \ \ \ \ \ P(x)\ \ \ \ \ \ \ x\times P(x)\ \ \ \ \ \ \ (x-\mu)^2\times P(x)\\\\480\ \ \ \ \ \ \ 0.02\ \ \ \ \ \ \ 9.6\ \ \ \ \ \ \ 72.9632\\\\490\ \ \ \ \ \ \ 0.05\ \ \ \ \ \ \ 24.5\ \ \ \ \ \ \ 127.008\\\\500\ \ \ \ \ \ \ 0.14\ \ \ \ \ \ \ 70\ \ \ \ \ \ \ 228.5024\\\\510\ \ \ \ \ \ \ 0.16\ \ \ \ \ \ \ 81.6\ \ \ \ \ \ \ 147.8656\\\\520\ \ \ \ \ \ \ 0.14\ \ \ \ \ \ \ 72.8\ \ \ \ \ \ \ 58.2624\\\\550\ \ \ \ \ \ \ 0.18\ \ \ \ \ \ \ 99\ \ \ \ \ \ \ 16.5888$

$580\ \ \ \ \ \ \ 0.18\ \ \ \ \ \ \ 104.4\ \ \ \ \ \ \ 282.2688\\\\600\ \ \ \ \ \ \ 0.12\ \ \ \ \ \ \ 72\ \ \ \ \ \ \ 426.2592\\\\650\ \ \ \ \ \ \ 0.01\ \ \ \ \ \ \ 6.5\ \ \ \ \ \ \ 120.1216$

            $Total\ \ \ \ \ \ \ \ \ \ 540.4 \ \ \ \ \ \ \ \ \ \ 1479.84\\\\Mean \ \ \ \ \ \ \ \ \ \ 540.4 \ \ \ \ \ \ \ \ \ \ dollars\\\\Variance \ \ \ \ \ \ \ 1479.84 \ \ \ \ \ \ \ \ dollars^2\\\\St \ Dev \ \ \ \ \ \ \ \ \ \ 38 \ \ \ \ \ \ \ \ \ \ dollars$