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.

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 and standard deviation , the zscore of a measure X is given by:

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:

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

has a pvalue of 0.

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

15.5

step-by-step explanation:

5

∑

n = 1

substitute n = 1, 2, 3, 4, 5 and sum the terms

= + + + 2² + 2³

= 0.5 + 1 + 2 + 4 + 8

= 15.5