The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean =41 and standard deviation =6
(a) What is the probability that a randomly chosen tire has a lifetime greater than 50 thousand miles?

6.68% probability that a randomly chosen tire has a lifetime greater than 50 thousand miles

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:

(a) What is the probability that a randomly chosen tire has a lifetime greater than 50 thousand miles?

This is 1 subtracted by the pvalue of Z when X = 50. So

has a pvalue of 0.9332

1 - 0.9332 = 0.0668

6.68% probability that a randomly chosen tire has a lifetime greater than 50 thousand miles

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.