Respuesta :
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([tex]\mu=150,000 ,\sigma=2,300[/tex])
The z score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean price = $150,000
[tex]\sigma[/tex] = standard deviation = $2,300
Now, according to 68-95-99.7 rule;
Around 68% of the values in a normal distribution lies between [tex]\mu-\sigma[/tex] and [tex]\mu-\sigma[/tex].
Around 95% of the values occur between [tex]\mu-2\sigma[/tex] and [tex]\mu+2\sigma[/tex] .
Around 99.7% of the values occur between [tex]\mu-3\sigma[/tex] and [tex]\mu+3\sigma[/tex].
So, firstly we will find the z scores for both the values given;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] = [tex]\frac{147,700-150,000}{2,300}[/tex] = -1
Z = [tex]\frac{X-\mu}{\sigma}[/tex] = [tex]\frac{152,300-150,000}{2,300}[/tex] = 1
This indicates that we are in the category of between [tex]\mu-\sigma[/tex] and [tex]\mu-\sigma[/tex].
SO, this represents that percentage of buyers who paid between $147,700 and $152,300 is 68%.
