Given:
Mean,
[tex]\mu = 300,000[/tex]
Standard deviation,
[tex]\sigma = 50,000[/tex]
The Chebyshev theorem will be used for the solution of the given query,
i.e., [tex]P(|X - \mu| < k \sigma ) \geq 1-\frac{1}{k^2}[/tex]
(i)
⇒ [tex]1-\frac{1}{k^2} = 0.75[/tex]
[tex]k = 2[/tex]
The lower limit will be:
= [tex]\mu - k \sigma[/tex]
= [tex]300000-2\times 50000[/tex]
= [tex]200000[/tex]
The upper limit will be:
= [tex]\mu + k \sigma[/tex]
= [tex]300000+2\times 50000[/tex]
= [tex]400000[/tex]
hence,
The 75% houses sold are between:
⇒ (200000, 400000)
(ii)
⇒ [tex]\mu -k \sigma = Lower \ limit[/tex]
By putting the values, we get
[tex]300000-50000k = 150000[/tex]
[tex]300000-150000=50000k[/tex]
[tex]150000=50000k[/tex]
[tex]k=\frac{150000}{50000}[/tex]
[tex]=3[/tex]
now,
⇒ [tex]1-\frac{1}{k^2} = 1-\frac{1}{3^2}[/tex]
[tex]=\frac{8}{9}[/tex]
[tex]=0.8888[/tex]
[tex]=88.88[/tex] (%)
hence,
88.88% houses sold between 150000 and 450000.
(iii)
⇒ [tex]\mu - k \sigma = Lower \ limit[/tex]
By putting the values, we get
[tex]300000-50000k = 170000[/tex]
[tex]300000-170000=50000 k[/tex]
[tex]130000= 50000 k[/tex]
[tex]k = \frac{130000}{50000}[/tex]
[tex]=2.6[/tex]
now,
⇒ [tex]1-\frac{1}{k^2} = 1-\frac{1}{2.6^2}[/tex]
[tex]=0.8521[/tex]
[tex]=85.21[/tex] (%)
hence,
85.21% houses are sold between 170000 and 430000.
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