Question

Suppose the distribution of home sales prices has mean k300, 000 and standard

Answer 1

Given:

Mean,

$\mu = 300,000$

Standard deviation,

$\sigma = 50,000$

The Chebyshev theorem will be used for the solution of the given query,

i.e., $P(|X - \mu| < k \sigma ) \geq 1-\frac{1}{k^2}$

(i)

⇒ $1-\frac{1}{k^2} = 0.75$

            $k = 2$

The lower limit will be:

= $\mu - k \sigma$

= $300000-2\times 50000$

= $200000$

The upper limit will be:

= $\mu + k \sigma$

= $300000+2\times 50000$

= $400000$

hence,

The 75% houses sold are between:

⇒ (200000, 400000)

(ii)

⇒ $\mu -k \sigma = Lower \ limit$

By putting the values, we get

   $300000-50000k = 150000$

   $300000-150000=50000k$

                  $150000=50000k$

                           $k=\frac{150000}{50000}$

                              $=3$

now,

⇒ $1-\frac{1}{k^2} = 1-\frac{1}{3^2}$

              $=\frac{8}{9}$

              $=0.8888$

              $=88.88$ (%)

hence,

88.88% houses sold between 150000 and 450000.      

(iii)

⇒ $\mu - k \sigma = Lower \ limit$

By putting the values, we get

   $300000-50000k = 170000$

   $300000-170000=50000 k$

                  $130000= 50000 k$

                          $k = \frac{130000}{50000}$

                             $=2.6$

now,

⇒ $1-\frac{1}{k^2} = 1-\frac{1}{2.6^2}$

              $=0.8521$

              $=85.21$ (%)

hence,

85.21% houses are sold between 170000 and 430000.

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