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Math Help - distrubition

  1. #1
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    distrubition

    I'm working through my review sheet before my final Friday and I have questions. I think I got the first right (check my work?) but the second I'm lost on

    "Suppose housing prices observe a normal distribution. For the Pleasantville housing market, the average housing price is $225,000 and the standard deviation is $40100. What percent of houses in Pleasantville are more expensive than $300,000?"

    Z= (X- pop mean) / st. dev.
    Z= (300 ,000-225,000) / 40,100= 1.87
    P(0<z<1.87)= 0.4692
    0.5-0.4692= 0.0308.
    So, 3.1% of houses in Pleasantville are more expensive than $300,000.


    "Suppose that 3.5% of the houses in the area get a property tax exemption due to their low values. Find the value above which households are not qualified for the exemption?"

    I don't even know how to begin this one, am I looking for the value above which the *lowest* 3.5% are? 0.50-0.035=0.465?
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by peanutbutter View Post
    I'm working through my review sheet before my final Friday and I have questions. I think I got the first right (check my work?) but the second I'm lost on

    "Suppose housing prices observe a normal distribution. For the Pleasantville housing market, the average housing price is $225,000 and the standard deviation is $40100. What percent of houses in Pleasantville are more expensive than $300,000?"

    Z= (X- pop mean) / st. dev.
    Z= (300 ,000-225,000) / 40,100= 1.87
    P(0<z<1.87)= 0.4692
    0.5-0.4692= 0.0308.
    So, 3.1% of houses in Pleasantville are more expensive than $300,000.


    "Suppose that 3.5% of the houses in the area get a property tax exemption due to their low values. Find the value above which households are not qualified for the exemption?"

    I don't even know how to begin this one, am I looking for the value above which the *lowest* 3.5% are? 0.50-0.035=0.465?
    P(X > 300k) \\ = P \left( \frac{X - 225k}{40.1k} > \frac{300k - 225k}{40.1k} \right)

    = P(Z > 1.87032)

    = 1 - P(Z < 1.87032)

    = 1 - \Phi (1.87032)

    = 1 - 0.9693

    = 0,0307

    Roughly 3,07% of the houses are priced over $300 000


    =================


    "Suppose that 3.5% of the houses in the area get a property tax exemption due to their low values. Find the value above which households are not qualified for the exemption?"

    P(X < Price) \\ = P \left( \frac{X - 225k}{40.1k} < \frac{Price - 225k}{40.1k} \right) = 0,035

    P \left( Z < \frac{Price - 225k}{40.1k} \right)  = 0,035

    \Phi \left(\frac{Price - 225k}{40.1k} \right)  = 0,035

    So \Phi of a certain value must give you 0,035
    To get that value, use your standard normal table in "reverse".

    My table is more inaccurate since it has less decimal places than the average table. So recheck it.

    \Phi \left( - \frac{Price - 225k}{40.1k} \right)  = 0,965

    - \frac{Price - 225k}{40.1k} = 1.81

    Price = 152k \ approx.
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