Question

e The Graph Illustrates A Normal[ × MMyOpenMath Assessment O (8) How To Take AScreenshot0 + × × (--) С https://www.myopenmath.com/assessment/showtest.php The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean pric 56:04 remaining. paid is $1800 and the standard deviation is $75. Questions Q4 Print Version 1575 1650 1725 1800 1875 1950 2025 Distribution of Prices What is the approximate percentage of buyers who paid between S1800 and S1875? What is the approximate percentage of buyers who paid more than $2025? What is the approximate percentage of buyers who paid between $1575 and $1800? What is the approximate percentage of buyers who paid between $1725 and $1875? What is the approximate percen han $1650? 13:41 Lu b pai

%

can you explain how to do this for me

Answer 1

Solution:-

Mean = 1800, S.D = 75

a) The percentage of buyers who paid between $1800 and $1875 is 34.13%.

x₁ = 1800

x₂ = 1875

By applying normal distribution:-

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

z₁ = 0

z₂ = 1.0

P( 0 < z < 1.0) = P(z > 0) - P(z > 1.0)

P( 0 < z < 1.0) = 0.50 - 0.1587

P( 0 < z < 1.0) = 0.3413

b) The approximate percentage of buyers who paid more than $2025 is 0.13%.

x = 2025

By applying normal distribution:-

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

z = 3.0

P(z > 3.0) = 0.0013

c) The percentage of buyers who paid between $1575 and $1800 is 49.87%.

x₁ = 1575

x₂ = 1800

By applying normal distribution:-

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

z₁ = -3.0

z₂ = 0

P( -3.0 < z < 0) = P(z > -3.0) - P(z > 0)

P( -3.0 < z < 0) = 0.9987 - 0.50

P( -3.0 < z < 0) = 0.4987

d) The percentage of buyers who paid between $1725 and $1875 is 68.26%.

x₁ = 1725

x₂ = 1875

By applying normal distribution:-

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

z₁ = -1.0

z₂ = 1.0

P( -1.0 < z < 1.0) = P(z > -1.0) - P(z > 1.0)

P( -1.0 < z < 1.0) = 0.8413 - 0.1587

P( -1.0 < z < 1.0) = 0.6826

e) The percentage of buyers who paid less than $1650 is 2.28%.

x = 1650

By applying normal distribution:-

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

z = -2.0

P(z < -2.0) = 0.0228

f) The percentage of buyers who paid between $1800 and $1950 is 47.72%.

x₁ = 1800

x₂ = 1950

By applying normal distribution:-

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

z₁ = 0

z₂ = 2.0

P( 0 < z < 2.0) = P(z > 0) - P(z > 2.0)

P( 0 < z < 2.0) = 0.50 - 0.0228

P( 0 < z < 2.0) = 0.4772