a. 68 % between $147,700 and $152,300 if the standard deviation is $2300
b. 2.5 % more than $154,800 if the standard deviation is $2400
Given that:
Prices of a certain model of a new home are normally distributed with a mean of $150,000.
Using the 68-95-99.7 rule , 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 ( μ = 150,000 , σ = 2,300 )
z score probability distribution for normal distribution
Z = X - μ / σ
where μ population mean price = $150,000
σ standard deviation = $2,300
According to 68-95-99.7 rule;
Around 68% of the values in a normal distribution lies between μ - σ and μ +σ .
Around 95% of the values occur between μ - 2σ and μ + 2σ .
Around 99.7% of the values occur between μ - 3σ and μ + 3σ.
Find the z scores for both given values
Z = X - μ / σ
= [tex]\frac{147,700 - 150,000}{2,300}\\ \\ \\[/tex]
= -1
Z = X - μ / σ
= [tex]\frac{152,300 - 150,000}{2,300}[/tex]
= -1
The z score indicates that the category of between μ - σ and μ +σ .
Therefore , this represents that percentage of buyers who paid between $147,700 and $152,300 is 68%.
To learn more about normal distribution check the given link
https://brainly.com/question/25224028
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