Respuesta :
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.
[tex]z = \dfrac{X - \mu}{\sigma}[/tex]
For the sample mean x = 150000
[tex]z = \dfrac{150000 - 150000}{1200}[/tex]
[tex]z = \dfrac{0}{1200}[/tex]
z = 0
For the sample mean x = 152400
[tex]z = \dfrac{152400 - 150000}{1200}[/tex]
[tex]z = \dfrac{2400 }{1200}[/tex]
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
This question is based on concept of statistics. Therefore, correct option is C i.e. 47.5% of buyers who paid: between $150,000 and $152,400 if the standard deviation is $1200.
Given:
Mean is $150,000, and
Standard deviation is $1200.
We need to determined the percentage of buyers who paid: between $150,000 and $152,400 as per given mean and standard deviation.
By using z score formula can be use to calculate the percentage of the buyer who paid,
[tex]\bold{z=\dfrac{x-\mu }{\sigma}}[/tex]
As given in question sample mean i.e. X= 150,000
[tex]z=\dfrac{150000-150000}{1200} \\\\z= \dfrac{0}{1200}\\\\z=0[/tex]
Now for the sample mean X = 152,400 ,
[tex]z=\dfrac{152400-150000}{1200} \\\\\\z= \dfrac{24000}{1200}\\\\\\z=2[/tex]
By using standard normal distribution table,
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, correct option is C that is 47.5%.
For further details, please prefer this link:
https://brainly.com/question/23907081
