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 $150,000 and $152,400 if the standard deviation is $1200.
A. 68%
B. 99.7%
C. 47.5%
D. 34%

Answer 1

Answer:

C. 47.5%

Step-by-step explanation:

The  summary of the given statistics include:

mean =150000

standard deviation: 1200

The  objective is to use tributed with a mean of $150,000. Use the 68-95-99.7 rule to find the percentage of buyers who paid: between $150,000 and $152,400

The z score formula can be use to calculate the percentage of the buyer who paid.

$z = \dfrac{X - \mu}{\sigma}$

For the sample mean x = 150000

$z = \dfrac{150000 - 150000}{1200}$

$z = \dfrac{0}{1200}$

z = 0

For the sample mean x = 152400

$z = \dfrac{152400 - 150000}{1200}$

$z = \dfrac{2400 }{1200}$

z  = 2

From the standard normal distribution tables

P(150000 < X < 152400) = P(0 < z < 2 )

P(150000 < X < 152400) =P(z<2) -P(z<0)

P(150000 < X < 152400) =0.9772 -0.5

P(150000 < X < 152400) = 0.4772

P(150000 < X < 152400) = 47.7%  which is close to 47.5% therefore option C is correct