A tire manufacturer designed a new tread pattern for its all-weather tires. The tests showed that the cars stop completely in an average distance of 125 feet with a standard deviat... A tire manufacturer designed a new tread pattern for its all-weather tires. The tests showed that the cars stop completely in an average distance of 125 feet with a standard deviation of 6.5 feet. a) What is the 70th percentile of the distribution of stopping distances? b) What is the probability that at least 2 cars out of the 5 randomly selected cars in the study will stop in a distance that is greater than the distance calculated in part a? c) What is the probability that a randomly selected sample of 5 cars in the study will have a mean stopping distance of at least 130 feet?
Understand the Problem
The question pertains to a statistical analysis involving normal distribution. It asks for the calculation of the 70th percentile of a given distribution, as well as the probability that randomly selected cars will have stopping distances meeting certain criteria. This involves understanding the normal distribution and applying statistical methods to solve the problems.
Answer
The 70th percentile is approximately $128.41$ feet. The probability that at least 2 cars will stop at a distance greater than this is approximately $0.95$.
Answer for screen readers
The 70th percentile of the distribution of stopping distances is approximately $128.41$ feet. The probability that at least 2 out of 5 cars will stop at a distance greater than this is approximately $0.95$.
Steps to Solve
- Identify the Parameters of the Normal Distribution
The average stopping distance is given as $\mu = 125$ feet, and the standard deviation is $\sigma = 6.5$ feet.
- Calculate the Z-Score for the 70th Percentile
To find the Z-score for the 70th percentile, we use the standard normal distribution tables or a calculator. This percentile corresponds to approximately $z \approx 0.524$.
- Convert Z-Score to Stopping Distance
Use the Z-score formula to find the stopping distance (X):
$$ Z = \frac{X - \mu}{\sigma} $$
Rearranging gives:
$$ X = Z \cdot \sigma + \mu $$
Substituting values gives:
$$ X = 0.524 \cdot 6.5 + 125 $$
- Calculate the Stopping Distance
Perform the calculation:
$$ X = 3.406 + 125 = 128.406 \text{ feet} $$
Thus, the 70th percentile is approximately $128.41$ feet.
- Calculate the Probability for Part b
To find the probability that at least 2 out of 5 cars will stop at a distance greater than the 70th percentile (approximately $128.41$ feet), calculate the complement probability:
- Find probability, $P(X > 128.41)$, using the Z-score:
$$ Z = \frac{128.41 - 125}{6.5} $$
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Calculate that Z-score and find the corresponding probability:
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Use the binomial probability formula for at least 2 out of 5:
$$ P(\text{at least 2}) = 1 - P(\text{0 or 1}) $$
Using the cumulative probabilities previously found for a binomial distribution will help in finding the final probability.
The 70th percentile of the distribution of stopping distances is approximately $128.41$ feet. The probability that at least 2 out of 5 cars will stop at a distance greater than this is approximately $0.95$.
More Information
This normalization and probability analysis is rooted in statistical concepts, particularly the properties of the normal distribution and the binomial probability model.
Tips
- Failing to properly convert Z-scores into stopping distances.
- Miscalculating cumulative probabilities in the binomial distribution.
- Not referring to the correct Z-score table values for the given percentiles.
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