Respuesta :
Answer:
(a) The probability that for a sample of 35 individuals that purchase a used car will pay an average between $6820 to $7880 is 0.9120.
(b) The probability that for a sample of 35 individuals that purchase a used car will pay an average of more than $ 7140 is 0.7258.
Step-by-step explanation:
Let X = amount of money individuals pay when buying a used car.
The random variable X is Normally distributed with mean μ = $7310 and standard deviation σ = $1640.
A sample of n = 35 individuals who purchase a used car is selected.
We need to compute the probability of:
(a) Between $6820 to $7880
(b) More then $7140.
(a)
Compute the probability that for a sample of 35 individuals that purchase a used car will pay an average between $6820 to $7880 as follows:
[tex]P(6820<\bar X<7880)=P(\frac{6820-7310}{1640/\sqrt{35}}<\frac{\bar X-\mu}{\sigma/\sqrt{n}}<\frac{7880-7310}{1640/\sqrt{35}})[/tex]
[tex]=P(-1.77<Z<2.06)\\=P(Z<2.06)-P(Z<-1.77)\\=0.98030-0.03836\\=0.94194\\\approx 0.9120[/tex]
*Use a z-table for the probability.
Thus, the probability that for a sample of 35 individuals that purchase a used car will pay an average between $6820 to $7880 is 0.9120.
(b)
Compute the probability that for a sample of 35 individuals that purchase a used car will pay an average of more than $ 7140 as follows:
[tex]P(\bar X>7140)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{7140-7310}{1640/\sqrt{35}})[/tex]
[tex]=P(Z>-0.61)\\=P(Z<0.61)\\=0.72575\\\approx0.7258[/tex]
*Use a z-table for the probability.
Thus, the probability that for a sample of 35 individuals that purchase a used car will pay an average of more than $ 7140 is 0.7258.
