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AleksandrR [38]
3 years ago
8

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

Mathematics
1 answer:
777dan777 [17]3 years ago
3 0

<u>Given</u>:

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.

Learn more:

brainly.com/question/20897803      

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