Respuesta :
Answer:
The longest braking distance one of these cars could have and still be in the bottom 1% is of 116.94 feet.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
The braking distances of a sample of cars are normally distributed, with a mean of 129 feet and a standard deviation of 5.18 feet.
This means that [tex]\mu = 129, \sigma = 5.18[/tex]
What is the longest braking distance one of these cars could have and still be in the bottom 1%?
This is the 1st percentile, which is X when Z has a pvalue of 0.01, so X when Z = -2.327.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-2.327 = \frac{X - 129}{5.18}[/tex]
[tex]X - 129 = -2.327*5.18[/tex]
[tex]X = 116.94[/tex]
The longest braking distance one of these cars could have and still be in the bottom 1% is of 116.94 feet.
