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konstantin123 [22]
3 years ago
5

The mean price of a home in California is much larger than the median price. Explain why this might happen.

Mathematics
1 answer:
vivado [14]3 years ago
8 0
The median price is that which is midway in the range between the cheapest and the dearest. 
<span>The mean price is the sum of all the prices divided by the number </span>
<span>For example, lets say we have seven houses in a particular road. I haven't a clue how much Californian houses cost, so I'm going to list seven arbitrary values in order of cheapest to dearest. </span>

<span>1) $100,000 </span>
<span>2) $130,000 </span>
<span>3) $130,000 </span>
<span>4) $250,000 </span>
<span>5) $310,000 </span>
<span>6) $800,000 </span>
<span>7) $900,000 </span>

<span>If the size n, of the range is odd, then the median value is at the (n+1)/2 th position in the range </span>
<span>For n = 7, (n+1)/2 = (7+1)/2 = 8/2 = 4 </span>
<span>The fourth cheapest house is priced at $250,000, which is the median </span>

<span>The sum of the cost of all seven houses is $2,620,000 </span>
<span>The mean = Sum/n </span>
<span>= $2,620,000 / 7 </span>
<span>= $374,285.71 </span>

<span>In summary </span>
<span>Median = $250,000 </span>
<span>Mean = $374,285.71 </span>

<span>The reason why the mean is much greater than the median is because of the skewed values at the top of the range. They are much higher than the others. If the prices of the sixth and seventh houses were $350,000 and $360,000 respectively, the median would still be $250,000, but now the mean would be $232,857.14 which is less than the median. </span>

<span>Unlike the median, the mean is very sensitive to the extreme points within the range. </span>

<span>If it's a single home, or all homes are the same price, then the mean and the median are equal.</span>
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μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

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Step-by-step explanation:

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<em>Distance 1:</em>

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Kf = 0.5*m*vf² = 0.5*m*v²

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then

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⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

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