Answer:
(a) 0.0051
(b) 0.9773
Step-by-step explanation:
Let X = prices of the car.
It is provided that [tex]X\sim N(23,000, 3,500^{2})[/tex].
(a)
Compute the value of P (X ≥ 32000) as follows:
[tex]P(X\geq 32000)=P(\frac{X-\mu}{\sigma}>\frac{32000-23000}{3500})[/tex]
[tex]=P(Z>2.57)\\=1-P(Z<2.57)\\=1-0.99492\\=0.00508\\\approx 0.0051[/tex]
Thus, the proportion of cars cost an equal amount or more than $32,000 is 0.0051.
(b)
Compute the value of P (X ≥ 16000) as follows:
[tex]P(X\geq 16000)=P(\frac{X-\mu}{\sigma}>\frac{16000-23000}{3500})[/tex]
[tex]=P(Z>-2)\\=P(Z<2)\\=0.97725\\\approx 0.9773[/tex]
Thus, the proportion of cars cost an equal amount or more than $16,000 is 0.9773.