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The problem

"The mean value of houses on a street is \S125,000 with a standard deviation of \$5,000. The dataset has a mound shaped distribution.

(a) Estimate the percent of houses between \$120,000 and \$135,000.

(b) Estimate the percent of houses below \$115,000."

I was given this homework problem on my Statistics one homework and I don't know how to solve this. The book doesn't have any examples of this type of problem. Any assistance would be appreciated.

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  • $\begingroup$ Welcome to the site. Please take "the tour" to learn how to get best responses. Please repeat the question rather than giving a link (which may not be permanent). If you need to type \$ you have to use \$ because we use \$ for mathematical expressions. Also, please say what you have tried and why you can't finish the problem. That way we'll know how to address your difficulty. $\endgroup$ – BruceET Sep 11 '18 at 0:44
  • $\begingroup$ Look up 'standardization' or z-scores in your text. Standard scores enable you to use standard normal tables. Look at a couple of examples in your text how to use printed normal tables (or software). // In (a), raw score \$120,000 score corresponds to standard score $Z = \frac{120,000 - 125,000}{5000} = 1.$ Can you find the $Z$ score for raw score \$135,000 ? // Using software instead of printed normal tables in (a), I get about 82%. $\endgroup$ – BruceET Sep 11 '18 at 0:56
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I doubt very much that your stats book doesn't have information on the areas/probabilities for different regions of a normal distribution curve from around $-3$ to $+3$ standard deviations in single standard deviation regions. Nevertheless..........

Part a. is the region between one standard deviation below and two standard deviations above the mean. This is the area of the region between a Z score from $-1$ to $+2$ stated as a percentage of one.

Part b. is the area of region to the left of two standard deviation below the mean. This is the area below a Z score of $-2$ stated as a percentage of one.

Your book should have a table of Z sores. If not, use the table in this link http://users.stat.ufl.edu/~athienit/Tables/Ztable.pdf

For part a. The percentage of homes in this price range is the area to the left of a z score of $+2.0$ minus the area to the left for a Z score of $-1.0$ multiplied by $100$.

For part b. its simply the area to the left of a z score of $-2$ so simply read off the value in the table and multiply it by $100$ to get the percentage of homes in this price bracket.

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