The life expectancy of a particular brand of tire is normally distributed with a mean of 50,000 and a standard deviation of 4,000 miles.

The life expectancy of a particular brand of tire is normally distributed with a mean of 50,000 and a standard deviation of 4,000 miles. What percentage of tires will have a life of 42,600 to 57,400 miles? a. 46.78% b. 93.57% c. 50% d. 23.39%

The Correct Answer and Explanation is:

To solve this problem, we need to use the concept of the normal distribution and z-scores to find the percentage of tires that will have a life expectancy between 42,600 and 57,400 miles. The mean life expectancy of the tires is 50,000 miles, and the standard deviation is 4,000 miles.

Step 1: Standardize the Values

We first convert the given values (42,600 and 57,400 miles) into z-scores, which measure how many standard deviations each value is away from the mean. The formula for a z-score is: z=X−μσz = \frac{X – \mu}{\sigma}z=σX−μ​

Where:

  • XXX is the value we’re interested in,
  • μ\muμ is the mean,
  • σ\sigmaσ is the standard deviation.

For 42,600 miles:

z1=42,600−50,0004,000=−7,4004,000=−1.85z_1 = \frac{42,600 – 50,000}{4,000} = \frac{-7,400}{4,000} = -1.85z1​=4,00042,600−50,000​=4,000−7,400​=−1.85

For 57,400 miles:

z2=57,400−50,0004,000=7,4004,000=1.85z_2 = \frac{57,400 – 50,000}{4,000} = \frac{7,400}{4,000} = 1.85z2​=4,00057,400−50,000​=4,0007,400​=1.85

Step 2: Look Up the Z-Scores in the Standard Normal Distribution Table

Using the z-table, we find the probabilities corresponding to the z-scores of -1.85 and 1.85.

  • For z=−1.85z = -1.85z=−1.85, the cumulative probability is approximately 0.0322.
  • For z=1.85z = 1.85z=1.85, the cumulative probability is approximately 0.9678.

Step 3: Calculate the Percentage

To find the percentage of tires between 42,600 and 57,400 miles, we subtract the cumulative probability for z=−1.85z = -1.85z=−1.85 from the cumulative probability for z=1.85z = 1.85z=1.85: P(42,600≤X≤57,400)=P(z2)−P(z1)=0.9678−0.0322=0.9356P(42,600 \leq X \leq 57,400) = P(z_2) – P(z_1) = 0.9678 – 0.0322 = 0.9356P(42,600≤X≤57,400)=P(z2​)−P(z1​)=0.9678−0.0322=0.9356

Thus, approximately 93.57% of the tires will have a life expectancy between 42,600 and 57,400 miles.

Correct Answer: b. 93.57%

Conclusion:

By using z-scores and the cumulative probability from the normal distribution table, we can find that about 93.57% of the tires will last between 42,600 and 57,400 miles. This is a direct application of probability theory involving the normal distribution.

Scroll to Top