Respuesta :
Answer:
The percentage of recently sold homes with prices between $217,700 and $242,300
P( 2,17,700 ≤X≤2,42,300) = 95.44
Step-by-step explanation:
Step(i):-
Given that the mean the Population = 2,30,000
Given that the standard deviation of the Normal distribution = 6150
By using
[tex]Z = \frac{x-mean}{S.D}[/tex]
Let 'X₁' = 2,17,700
[tex]Z_{1} = \frac{217700-230000}{6150} = -2[/tex]
Let 'X₂' = 2,42,300
[tex]Z_{2} = \frac{242300-230000}{6150} = 2[/tex]
Step(ii):-
The probability that recently sold homes with prices between $217,700 and $242,300
P( 2,17,700 ≤X≤2,42,300) = P( -2≤Z≤2)
= | A(2) + A(-2)|
= |A(2) + A(2)| (∵A(-z) =A(z)
= 2× A(2)
= 2×0.4772
= 0.9544
Final answer:-
The probability that recently sold homes with prices between $217,700 and $242,300
P( 2,17,700 ≤X≤2,42,300) = 0.9544
The percentage of recently sold homes with prices between $217,700 and $242,300
P( 2,17,700 ≤X≤2,42,300) = 95.44
