Suppose that prices of a certain model of a new home are normally distributed with a mean of $150,000. Use the 68-95-99.7 rule to find the percentage of buyers who paid between $147,700 and $152,300 if the standard deviation is $2300.

68% of buyers paid between $147,700 and $152,300.

Step-by-step explanation:

We are given that prices of a certain model of a new home are normally distributed with a mean of $150,000.

Use the 68-95-99.7 rule to find the percentage of buyers who paid between $147,700 and $152,300 if the standard deviation is $2300.

Let X = prices of a certain model of a new home

SO, X ~ Normal()

The z score probability distribution for normal distribution is given by;

                             Z  =    ~ N(0,1)

where, = population mean price = $150,000

            = standard deviation = $2,300

Now, according to 68-95-99.7 rule;

Around 68% of the values in a normal distribution lies between and .

Around 95% of the values occur between and .

Around 99.7% of the values occur between and .

So, firstly we will find the z scores for both the values given;

         Z  =    =    = -1

         Z  =    =    = 1

This indicates that we are in the category of between and .

SO, this represents that percentage of buyers who paid between $147,700 and $152,300 is 68%.

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