Question

The average amount of money individuals pay when buying a used car is $7,310 and the standard deviation is $1,640. Assume the amount of money individuals pay when buying a used car is normally distributed. Find the probability that for a sample of 35 individuals that purchase a used car, they will pay an average (Show your work to receive credit)

Answer 1

(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.

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:

`P(6820<\bar X<7880)=P((6820-7310)/(1640/√(35))<(\bar X-\mu)/(\sigma/√(n))<(7880-7310)/(1640/√(35)))`

`=P(-1.77<Z<2.06)\\=P(Z<2.06)-P(Z<-1.77)\\=0.98030-0.03836\\=0.94194\\\approx 0.9120`

*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:

`P(\bar X>7140)=P((\bar X-\mu)/(\sigma/√(n))>(7140-7310)/(1640/√(35)))`

`=P(Z>-0.61)\\=P(Z<0.61)\\=0.72575\\\approx0.7258`

*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.