The average lifespan of a set of tires is 38,000 miles, with a standard deviation of 1500 miles. What is the probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

100% probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

Step-by-step explanation:

Problems of normally distributed samples are 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 that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

This is the pvalue of X when X = 44000 subtracted by the pvalue of Z when X = 32000. So

X = 44000

has a pvalue of 1.

X = 32000

has a pvalue of 0.

1 - 0 = 100%

100% probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

step-by-step explanation: