Suppose 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 p

Question

Suppose 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.

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Ariana 2 hours 2021-10-13T03:02:52+00:00 1 Answer 0

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    2021-10-13T03:03:56+00:00

    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(\mu=150,000 ,\sigma=2,300)

    The z score probability distribution for normal distribution is given by;

                                 Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

    where, \mu = population mean price = $150,000

                \sigma = standard deviation = $2,300

    Now, according to 68-95-99.7 rule;

    Around 68% of the values in a normal distribution lies between \mu-\sigma and \mu-\sigma.

    Around 95% of the values occur between \mu-2\sigma and \mu+2\sigma
    .

    Around 99.7% of the values occur between \mu-3\sigma and \mu+3\sigma.

    So, firstly we will find the z scores for both the values given;

             Z  =  \frac{X-\mu}{\sigma}  =  \frac{147,700-150,000}{2,300}  = -1

             Z  =  \frac{X-\mu}{\sigma}  =  \frac{152,300-150,000}{2,300}  = 1

    This indicates that we are in the category of between \mu-\sigma and \mu-\sigma.

    SO, this represents that percentage of buyers who paid between $147,700 and $152,300 is 68%.

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