Question

The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000 miles and a standard deviation of 1500 miles. What is the probability a certain tire of this brand will last between 56,850 miles and 57,300 miles

Answer 1

Answer:

0.018 is the required probability.              

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 60,000 miles

Standard Deviation, σ = 1500 miles

We are given that the distribution of tread life is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(brand will last between 56,850 miles and 57,300 miles)

$P(56850 \leq x \leq 57300) = P(\displaystyle\frac{56850 - 60000}{1500} \leq z \leq \displaystyle\frac{57300-60000}{1500}) = P(-2.1 \leq z \leq -1.8)\\\\= P(z \leq -1.8) - P(z < -2.1)\\= 0.0359 - 0.0179 = 0.018= 1.8\%$

$P(56850 \leq x \leq 57300) = 1.8\%$

0.018 is the probability a certain tire of this brand will last between 56,850 miles and 57,300 miles.