Question

A large sample of tires from cabs driven within a city have an average tire tread depth of 0.25cm at the end of the winter. If the city’s average winter tire tread depth is 2.2cm, with a standard deviation of 0.33cm at the end of the same winter, what is the probability cab tires depths would be shallower than 0.25cm.

Answer 1

Answer:

0% probability cab tires depths would be shallower than 0.25cm.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 2.2, \sigma = 0.33$

What is the probability cab tires depths would be shallower than 0.25cm.

This probability is the pvalue of Z when X = 0.25. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.25 - 2.2}{0.33}$

$Z = -5.9$

$Z = -5.9$ has a pvalue of 0.

0% probability cab tires depths would be shallower than 0.25cm.