Question

problem 9
the average price of a car is $25,500. assume the variable is normally distributed. if the standard deviation is $3,400, find the probability that a cars price will be:
a. less than $29,000
b. more than $31,000

Explanation:

Step1: Calculate the z - score for part a

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
For \(x = 29000\), \(\mu=25500\), \(\sigma = 3400\)
\(z=\frac{29000 - 25500}{3400}=\frac{3500}{3400}\approx1.03\)

Step2: Find the probability for part a

Using the standard normal distribution table, \(P(Z<1.03)\)
Looking up the value in the table, \(P(Z < 1.03)=0.8485\)

Step3: Calculate the z - score for part b

For \(x = 31000\), \(\mu = 25500\), \(\sigma=3400\)
\(z=\frac{31000 - 25500}{3400}=\frac{5500}{3400}\approx1.62\)

Step4: Find the probability for part b

\(P(X>31000)=1 - P(X\leq31000)\)
Since \(P(X\leq31000)=P(Z\leq1.62)\)
Looking up \(P(Z\leq1.62)\) in the standard - normal table, \(P(Z\leq1.62) = 0.9474\)
\(P(X>31000)=1 - 0.9474=0.0526\)

Answer:

a. The probability that a car's price is less than \(\$29000\) is \(0.8485\)
b. The probability that a car's price is more than \(\$31000\) is \(0.0526\)