Question

Suppose that prices of a certain model of new homes 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:

We want to find the percentage of values between 147700 and 152300

$P(147700$

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

$z= \frac{x-\mu}{\sigma}$

And replacing we got:

$z=\frac{147700-150000}{2300}=-1$

$z=\frac{152300-150000}{2300}=1$

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

Step-by-step explanation:

We define the random variable representing the prices of a certain model as X and the distirbution for this random variable is given by:

$X \sim N(\mu = 150000, \sigma =2300$

The empirical rule states that within one deviation from the mean we have 68% of the data, within 2 deviations from the mean we have 95% and within 3 deviations 99.7 % of the data.

We want to find the percentage of values between 147700 and 152300

$P(147700$

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

$z= \frac{x-\mu}{\sigma}$

And replacing we got:

$z=\frac{147700-150000}{2300}=-1$

$z=\frac{152300-150000}{2300}=1$

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%