Question

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.

Answer 1

Answer:

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($\mu=150,000 ,\sigma=2,300$)

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

                             Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = population mean price = $150,000

            $\sigma$ = standard deviation = $2,300

Now, according to 68-95-99.7 rule;

Around 68% of the values in a normal distribution lies between $\mu-\sigma$ and $\mu-\sigma$.

Around 95% of the values occur between $\mu-2\sigma$ and $\mu+2\sigma$ .

Around 99.7% of the values occur between $\mu-3\sigma$ and $\mu+3\sigma$.

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

         Z  =  $\frac{X-\mu}{\sigma}$  =  $\frac{147,700-150,000}{2,300}$  = -1

         Z  =  $\frac{X-\mu}{\sigma}$  =  $\frac{152,300-150,000}{2,300}$  = 1

This indicates that we are in the category of between $\mu-\sigma$ and $\mu-\sigma$.

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