Question

a tire manufacturer believes that the treadle of its snow tires can be described by a normal model with a mean of 22000 miles and standard deviation of 2700 miles. if you buy a set of these tires, what proportion of tires will last between 18000 and 26000 miles? which graph corresponds to the above proportion?

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x - \mu}{\sigma}\), where \(\mu = 22000\) (mean) and \(\sigma=2700\) (standard deviation).
For \(x = 18000\):
\(z_1=\frac{18000 - 22000}{2700}=\frac{- 4000}{2700}\approx - 1.48\)
For \(x = 26000\):
\(z_2=\frac{26000 - 22000}{2700}=\frac{4000}{2700}\approx1.48\)

Step2: Find probabilities

Using the standard normal table (or a calculator with a normal - distribution function), we know that \(P(Z < - 1.48)\) and \(P(Z>1.48)\) are related. Since the normal distribution is symmetric, \(P(Z < - 1.48)=P(Z>1.48)\) and \(P(-1.48 < Z < 1.48)=1 - 2P(Z < - 1.48)\)
From the standard normal table, \(P(Z < - 1.48)=0.0694\)
So \(P(-1.48 < Z < 1.48)=1-2\times0.0694 = 1 - 0.1388=0.8612\)

Answer:

The proportion of tires that will last between \(18000\) and \(26000\) miles is approximately \(0.8612\)