The tread life of tires mounted on light duty trucks follows the normal probability distribution with a mean of 60,000 miles and a standard deviation of 4,000 miles. suppose you bought a set of four tires, what is the likelihood the mean tire life of these four tires is more than 66,000 miles?

 0.0013

Step-by-step explanation:

Given : The test scores are normally distributed with

Mean :

Standard deviation :

Sample size :

The formula to calculate the z-score :-

For x = 66,000

The p-value =

Hence, the likelihood the mean tire life of these four tires is more than 66,000 miles = 0.0013

b

step-by-step explanation:

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3